Contents

Basic integrals 5

∫ dx, ∫ x^{a} dx, ∫ \frac{dx}{x}, ∫ \frac{f′(x)}{f(x)} dx, ∫ e^{ax} dx, ∫ e^{x} dx, ∫ a^{x} dx, ∫ sin ax dx, ∫ sin x dx, ∫ cos ax dx, ∫ cos x dx,

∫ \frac{dx}{cos^{2} ax}, ∫ \frac{dx}{cos^{2} x}, ∫ \frac{dx}{sin^{2} ax}, ∫ \frac{dx}{sin^{2} x}, ∫ \frac{dx}{\sqrt{a^{2} − x^{2}}} , ∫ \frac{dx}{\sqrt{x^{2} − a^{2}}} , ∫ \frac{dx}{\sqrt{x^{2} + a^{2}}} , ∫ \frac{dx}{x^{2} + a^{2}} , ∫ \frac{dx}{x^{2} − a^{2}} ,

∫ sinh ax dx, ∫ sinh x dx, ∫ cosh ax dx, ∫ cosh x dx, ∫ \frac{dx}{cosh^{2} ax}, ∫ \frac{dx}{cosh^{2} x}, ∫ \frac{dx}{sinh^{2} ax}, ∫ \frac{dx}{sinh^{2} x} . . . . . . . 5

Basic integrals — Appendix 6

∫ \frac{dx}{\sqrt{a^{2} − b^{2}x^{2}}} , ∫ \frac{dx}{\sqrt{b^{2}x^{2} − a^{2}}} , ∫ \frac{dx}{\sqrt{b^{2}x^{2} + a^{2}}} , ∫ \frac{dx}{b^{2}x^{2} + a^{2}} , ∫ \frac{dx}{b^{2}x^{2} − a^{2}} , ∫ \frac{dx}{a^{2} − b^{2}x^{2}} . . . . . . . . . . . . . . . . . . . 6

∫ \frac{dx}{\sqrt{x^{2} + qx + r}}, ∫ \frac{dx}{\sqrt{px^{2} + qx + r}} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

Substitution 8 ∫ \sqrt{1 − x^{2}} dx, ∫ sin^{3} t cos t dt, ∫ \frac{dx}{\sqrt{x + 1} + \sqrt[3]{x + 1}} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

∫ \frac{dx}{\sqrt{5 + 4 e^{x}}} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

∫ f (x, \sqrt[m]{\frac{ax + b}{cx + d}}) dx for ad − bc≠0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

∫ \frac{dx}{\sqrt{x + 1} + \sqrt[3]{x + 1}}, ∫ \begin{matrix}√ & \overline{1 − \sqrt{x}} \\ √ & \overline{1 + \sqrt{x}} \end{matrix} dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

∫ f(x, \sqrt{ax^{2} + bx + c}) dx for a≠0 [Euler substitutions] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

∫ \frac{dx}{\sqrt{x^{2} + a^{2}}} , ∫ \frac{dx}{\sqrt{a^{2} − x^{2}}} , ∫ \frac{dx}{\sqrt{x(1 − x)}} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

∫ f(sin x, cos x) dx [Universal trig substitution] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

∫ \frac{dx}{cos x}, ∫ \frac{dx}{1 + cos x}, ∫ \frac{dx}{sin x}, ∫ \frac{dx}{1 + sin x} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 Integration by parts (per partes) 13

∫ arctg x dx, ∫ x^{2} ln x dx, ∫ \frac{ln x}{\sqrt{x}} dx, ∫ x ln^{2} x dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 ∫ sin^{n} x dx, ∫ cos^{n} x dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

Partial fractions 15

∫ \frac{dx}{(x − a)^{n}}, ∫ \frac{dx}{x + 1}, ∫ \frac{1}{(x − 1)^{n}} dx, ∫ \frac{dx}{x^{2} + ax + b}, ∫ \frac{dx}{x^{2} + 4x + 25}, ∫ \frac{dx}{x^{2} + b}, ∫ \frac{dx}{x^{2} + β^{2}} , ∫ \frac{dx}{x^{2} + 1} . . . . . . . . . 15

∫ \frac{x + A}{(x^{2} + ax + b)^{n}} dx, ∫ \frac{2x + a}{(x^{2} + ax + b)^{n}} dx, ∫ \frac{dx}{(x^{2} + ax + b)^{n}} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

CONTENTS  Integrals

∫ \frac{dx}{(x^{2} + ax + b)^{2}} , ∫ \frac{dx}{(x^{2} + 4x + 25)^{2}} , ∫ \frac{dx}{(x^{2} + b)^{2}} , ∫ \frac{dx}{(x^{2} + β^{2})^{2}} , ∫ \frac{dx}{(x^{2} + 1)^{2}} , ∫ \frac{dx}{(x^{2} + ax + b)^{3}} , ∫ \frac{dx}{(x^{2} + 4x + 25)^{3}} ,

∫ \frac{dx}{(x^{2} + b)^{3}} , ∫ \frac{dx}{(x^{2} + β^{2})^{3}} , ∫ \frac{dx}{(x^{2} + 1)^{3}} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

∫ \frac{dx}{(x^{2} + ax + b)^{n}}, ∫ \frac{dx}{(x^{2} + b)^{n}}, ∫ \frac{dx}{(x^{2} + β^{2})^{n}}, ∫ \frac{dx}{(x^{2} + 1)^{n}}, ∫ \frac{x dx}{(x^{2} + 1)^{n}} . . . . . . . . . . . . . . . . . . . . . . . . . . 18

Fourier Series 19 Fourier Series of the function f(x), x ∈ ⟨a; a + 2π⟩ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 Fourier Series of the function f(x), x ∈ ⟨a; a + 2l⟩, l > 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 Fourier Sine Series and Fourier Cosine Series of the function f(x), x ∈ ⟨−π; π⟩ . . . . . . . . . . . . . . . . . . . . . . . 21 Fourier Series of the function f(x) = 1, x ∈ ⟨0 ; 2π⟩ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 Fourier Series of the function f(x) = 1, x ∈ ⟨−π; π⟩ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 Fourier Series of the function f(x) = 1, x ∈ ⟨0 ; π⟩ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 Fourier Series of the function f(x) = x, x ∈ ⟨0 ; 2π⟩ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 Fourier Series of the function f(x) = x, x ∈ ⟨0 ; π⟩ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 Fourier Series of the function f(x) = x, x ∈ ⟨−π; π⟩ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 Fourier Series of the function f(x) = x^{2}, x ∈ ⟨0 ; 2π⟩ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 Fourier Series of the function f(x) = x^{2}, x ∈ ⟨0 ; π⟩ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 Fourier Series of the function f(x) = x^{2}, x ∈ ⟨−π; π⟩ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

Integration and Series 32 ∫ e^{x} dx, ∫ e^{−x} dx, ∫ e^{(−x^{2})} dx, ∫ e^{(x^{2})} dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

∫ sin x dx, ∫ \frac{sin x}{x} dx, ∫ cos x dx, ∫ \frac{cos x}{x} dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

Examples 34

∫ [\sqrt{x^{3}} − \frac{1}{\sqrt{x}}] dx, ∫ x(x − a)(x − b) dx, ∫ e^{ax} dx, ∫ sinh^{2} x dx, ∫ cosh^{2} x dx, ∫ [tg x + cotg x] dx, ∫ \frac{x dx}{x^{2} − a^{2}} ,

∫ \frac{x dx}{x^{2} + a^{2}} , ∫ \frac{dx}{x}, ∫ \frac{x dx}{(x^{2} − a)^{n}}, ∫ \frac{ln cos x}{sin^{2} x} dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

∫ \frac{dx}{x^{6} + 1} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
∫ \frac{x dx}{x^{6} + 1} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
∫ \frac{x^{2} dx}{x^{6} + 1} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

∫ \frac{x^{3} dx}{x^{6} + 1}, ∫ \frac{x^{4} dx}{x^{6} + 1} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

∫ \frac{x^{5} dx}{x^{6} + 1}, ∫ \frac{x^{6} dx}{x^{6} + 1}, ∫ \frac{dx}{x ln x}, ∫ x^{2} ln x dx, ∫ \frac{ln x}{\sqrt{x}} dx, ∫ \frac{dx}{\sqrt{x(1 − x)}} . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

∫ \frac{dx}{(x − \sqrt{x^{2} − 1})^{2}} , ∫ \frac{1 + x}{\sqrt{1 − x^{2}}} dx, ∫ \frac{x^{n}}{x − 1} dx, ∫ \frac{x}{x − 1} dx, ∫ \frac{x^{3}}{x − 1} dx, ∫ \frac{x^{9}}{x − 1} dx . . . . . . . . . . . . . . . . . . . 41

∫ \frac{dx}{cos x}, ∫ \frac{dx}{1 + cos x}, ∫ \frac{dx}{sin x}, ∫ \frac{dx}{1 + sin x} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

∫ \frac{sin x dx}{\sqrt{cos^{5} x}}, ∫ \frac{cos x dx}{\sqrt[3]{sin^{2} x}}, ∫ sin^{2} x dx, ∫ cos^{2} x dx, ∫ sin^{3} x dx, ∫ cos^{3} x dx, ∫ sin^{n} x dx, ∫ sin^{2k+1} x dx, ∫ tg^{2} x dx …. 43

CONTENTS  Integrals

∫ cos^{n} x dx, ∫ cos^{2k+1} x dx, ∫ \frac{sin x − cos x}{\sqrt[4]{sin x + cos x}} dx, ∫ \frac{x^{2}}{sin x^{3}} dx, ∫ \frac{dx}{sin^{2} x cos^{2} x} . . . . . . . . . . . . . . . . . . . . . . . 44 ∫ cos ax sin bx dx, ∫ cos ax sin ax dx, ∫ x ln^{2} x dx, ∫ \frac{dx}{(1 − x)x^{2}} , ∫ ln x dx . . . . . . . . . . . . . . . . . . . . . . . . . 45 ∫ \sqrt{a^{2} − x^{2}} dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46 ∫ \sqrt{a^{2} + x^{2}} dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 ∫ \sqrt{x^{2} − a^{2}} dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 ∫ arctg x dx, ∫ x^{n} e^{x} dx, ∫ \frac{dx}{\sqrt{x + 1} + \sqrt[3]{x + 1}} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

∫ ln (1 + x^{2}) dx, ∫ \frac{dx}{\sqrt{x^{2} + a^{2}}} , ∫ \frac{dx}{\sqrt{x^{2} − a^{2}}} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

∫ \frac{dx}{\sqrt{a^{2} − x^{2}}} , ∫ \begin{matrix}√ & \overline{1 − \sqrt{x}} \\ √ & \overline{1 + \sqrt{x}} \end{matrix} dx, ∫ |x| dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

∫ x^{2} e^{2x} dx, ∫ x^{9} e^{2x} dx, ∫ \sqrt{\frac{1 + x}{1 − x}} dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

∫ \frac{dx}{x^{2}\sqrt{x^{2} − 1}}, ∫ x^{x}(ln x + 1) dx, ∫ ln (x + \sqrt{x^{2} + 1}) dx, ∫ \frac{dx}{\sqrt{x − 3} + \sqrt{x − 4}}, ∫ min_{x∈(0; ∞)} \{1, \frac{1}{x}\} dx . . . . . . . . . . . 54

∫ \frac{2x^{2} − x + 1}{x\sqrt{1 + x − x^{2}}} dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

∫ \frac{\sqrt{x^{2} − x}}{x} dx, ∫ \frac{x^{5} dx}{\sqrt{x^{3} + 1}} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

∫ \frac{dx}{\sqrt{5 + 4 e^{x}}}, ∫ \frac{\sqrt{1 − x^{2}}}{x^{2}} dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

∫ \sqrt{(\frac{1 − x}{1 + x})^{3}} dx, ∫ \sqrt{(\frac{1 + x}{1 − x})^{3}} dx, ∫ x^{2} ln \sqrt{1 − x} dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

∫ \frac{dx}{x^{6}(1 + x^{2})}, ∫ \sqrt{\frac{1 − x}{1 + x}} dx, ∫ \sqrt{\frac{1 + x}{1 − x}} dx, ∫ \sqrt{\frac{x − 1}{x + 1}} dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

∫ \sqrt{\frac{x + 1}{x − 1}} dx, ∫ \frac{dx}{x\sqrt{1 − x^{3} + x^{6}}} , ∫ \frac{dx}{x\sqrt{1 + x^{3} + x^{6}}} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

∫ ln (\sqrt{1 + x} + \sqrt{1 − x}) dx, ∫ \frac{1 − x}{x − \sqrt{x − x^{2}}} dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

∫ \frac{1 − x}{x − \sqrt{x^{2} − x}} dx, ∫ \frac{1 + \sqrt{1 − x^{2}}}{1 − \sqrt{1 − x^{2}}} dx, ∫ \sqrt{\frac{x}{1 − x\sqrt{x}}} dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

∫ \frac{dx}{\sqrt{e^{2x} + e^{x} +1}}, ∫ \sqrt{\frac{1 − e^{x}}{1 + e^{x}}} dx, ∫ \frac{1 − x}{x\sqrt{x − x^{2}}} dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

∫ \frac{dx}{a^{2} cos^{2} x + b^{2} sin^{2} x}, ∫ \frac{dx}{a^{2} cos^{2} x − b^{2} sin^{2} x}, ∫ x tg^{2} x dx, ∫ \frac{dx}{1 + \sqrt[3]{x}}, ∫ \frac{dx}{\sqrt[3]{x} + \sqrt[4]{x}} . . . . . . . . . . . . . . . . . . . . 64

∫ \frac{dx}{\sqrt[n]{x^{n} + 1}}, ∫ \frac{dx}{\sqrt[n]{1 − x^{n}}}, ∫ \frac{dx}{\sqrt[n]{x^{n} − 1}}, ∫ \frac{dx}{\sqrt{x^{2} + 1}} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

∫ \frac{dx}{\sqrt[3]{x^{3} + 1}} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

∫ \frac{dx}{\sqrt[4]{x^{4} + 1}}, ∫ \frac{dx}{\sqrt{x^{2} − 1}} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

\underline{Integrals}

∫ \frac{dx}{\sqrt[3]{x^{3} − 1}}, ∫ \frac{dx}{\sqrt[4]{x^{4} − 1}} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

∫ \frac{dx}{\sqrt{1 − x^{2}}} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

∫ \frac{dx}{\sqrt[3]{1 − x^{3}}} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

∫ \frac{dx}{\sqrt[4]{1 − x^{4}}} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

∫ sin nx dx, ∫ x sin nx dx, ∫ x^{2} sin nx dx, ∫ x^{3} sin nx dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

∫ cos nx dx, ∫ x cos nx dx, ∫ x^{2} cos nx dx, ∫ x^{3} cos nx dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

\underline{Integrals}

Basic integrals

∫ dx = ∫ 1 dx = x + c,  for x∈R (i.e. f(x) = 1, x∈R).  [c = constant]

∫ x^{a} dx = \frac{x^{a+1}}{a + 1} + c, for a∈R, a ≠ −1, x∈R − \{0\}.

∫ \frac{dx}{x} = ln |x| + c, for x∈R − \{0\}.

∫ \frac{f′(x)}{f(x)} dx = ln |f(x)| + c, for f(x)≠0, x∈D(f).

∫ e^{ax} dx = \frac{e^{ax}}{a} + c,  for a∈R, a ≠ 0, x∈R,  special case: ∫ e^{x} dx = e^{x} + c.
∫ a^{x} dx = \frac{a^{x}}{ln a} + c,  for a > 0, a ≠ 1, x∈R,  special case: ∫ e^{x} dx = \begin{matrix}e & x \\ ln & e \end{matrix} + c = e^{x} + c.
∫ sin ax dx = −\frac{cos ax}{a} + c,  for a∈R, a ≠ 0, x∈R,  special case: ∫ sin x dx = − cos x + c.
∫ cos ax dx = \frac{sin ax}{a} + c,  for a∈R, a ≠ 0, x∈R,  special case: ∫ cos x dx = sin x + c.
∫ \frac{dx}{cos^{2} ax} = \frac{tg ax}{a} + c,  for a∈R, a ≠ 0, x∈R − \{(2k + 1)\frac{π}{2} ; k ∈Z\},  special case: ∫ \frac{dx}{cos^{2} x} = tg x + c.
∫ \frac{dx}{sin^{2} ax} = −\frac{cotg ax}{a} + c,  for a∈R, a ≠ 0, x∈R − \{kπ; k ∈Z\},  special case: ∫ \frac{dx}{sin^{2} x} = − cotg x + c.

∫ \frac{dx}{\sqrt{a^{2} − x^{2}}} = \{ \begin{matrix}arcsin \frac{x}{|a|} + c, \\ − arccos \frac{x}{|a|} + c, \end{matrix} for a∈R, a ≠ 0, x∈(− |a|; |a|).

∫ \frac{dx}{\sqrt{x^{2} − a^{2}}} = ln |x + \sqrt{x^{2} − a^{2}}| + c, for a∈R, a ≠ 0, x∈(−∞; − |a|) ∪ (|a|; ∞).

∫ \frac{dx}{\sqrt{x^{2} + a^{2}}} = ln (x + \sqrt{x^{2} + a^{2}}) + c, for a∈R, a ≠ 0, x∈R.

∫ \frac{dx}{x^{2} + a^{2}} = \{ \begin{matrix}\frac{1}{a} arctg \frac{x}{a} + c, \\ −\frac{1}{a} arccotg \frac{x}{a} + c, \end{matrix} for a∈R, a ≠ 0, x∈R.

∫ \frac{dx}{x^{2} − a^{2}} = ∫ \frac{1}{2a} [\frac{1}{x − a} − \frac{1}{x + a}] dx = \frac{1}{2a} ln |\frac{x − a}{x + a}| + c, for a∈R, a ≠ 0, x∈R − \{±a\}.

∫ sinh ax dx = \frac{cosh ax}{a} + c,  for a∈R, a ≠ 0, x∈R,  special case: ∫ sinh x dx = cosh x + c.
∫ cosh ax dx = \frac{sinh ax}{a} + c,  for a∈R, a ≠ 0, x∈R,  special case: ∫ cosh x dx = sinh x + c.
∫ \frac{dx}{cosh^{2} ax} = \frac{tgh ax}{a} + c,  for a∈R, a ≠ 0, x∈R,  special case: ∫ \frac{dx}{cosh^{2} x} = tgh x + c.
∫ \frac{dx}{sinh^{2} ax} = −\frac{cotgh ax}{a} + c,  for a∈R, a ≠ 0, x∈R − \{0\},  special case: ∫ \frac{dx}{sinh^{2} x} = − cotgh x + c.

\underline{Integrals}

Basic integrals — Appendix

∫ \frac{dx}{\sqrt{a^{2} − b^{2}x^{2}}} = ∫ \frac{dx}{|b| \sqrt{\frac{a^{2}}{b^{2}} − x^{2}}} = \{ \begin{matrix}\frac{1}{|b|} arcsin \frac{|b| x}{|a|} + c, \\ −\frac{1}{|b|} arccos \frac{|b| x}{|a|} + c, \end{matrix} for a, b∈R − \{0\}, x∈(−\frac{|a|}{|b|} ; \frac{|a|}{|b|}).

∫ \frac{dx}{\sqrt{b^{2}x^{2} − a^{2}}} = ∫ \frac{dx}{|b| \sqrt{x^{2} − \frac{a^{2}}{b^{2}}}} = \frac{1}{|b|} ln |x + \sqrt{x^{2} − \frac{a^{2}}{b^{2}}} | + c_{1} = \frac{1}{|b|} ln |\frac{|b| x + \sqrt{b^{2}x^{2} − a^{2}}}{|b|} | + c_{1} =

= \frac{1}{|b|} ln ||b| x + \sqrt{b^{2}x^{2} − a^{2}}| − \frac{1}{|b|} ln |b| + c_{1} = \frac{1}{|b|} ln ||b| x + \sqrt{b^{2}x^{2} − a^{2}}| + c,

for a, b∈− \{0\}, x∈(−∞; −\frac{|a|}{|b|}) ∪ (\frac{|a|}{|b|} ; ∞).

∫ \frac{dx}{\sqrt{b^{2}x^{2} + a^{2}}} = ∫ \frac{dx}{|b| \sqrt{x^{2} + \frac{a^{2}}{b^{2}}}} = \frac{1}{|b|} ln (x + \sqrt{x^{2} + \frac{a^{2}}{b^{2}}} ) + c_{1} = \frac{1}{|b|} ln (\frac{|b| x + \sqrt{b^{2}x^{2} + a^{2}}}{|b|} ) + c_{1} =

= \frac{1}{|b|} ln (|b| x + \sqrt{b^{2}x^{2} + a^{2}}) − \frac{1}{|b|} ln |b| + c_{1} = \frac{1}{|b|} ln (|b| x + \sqrt{b^{2}x^{2} + a^{2}}) + c, for a, b∈R − \{0\}, x∈R.

∫ \frac{dx}{b^{2}x^{2} + a^{2}} = ∫ \frac{dx}{b^{2} (x^{2} + \frac{a^{2}}{b^{2}} )} = \{ \begin{matrix}\frac{1}{b^{2}} \frac{b}{a} arctg \frac{bx}{a} + c = \frac{1}{ab} arctg \frac{bx}{a} + c, \\ −\frac{1}{b^{2}} \frac{b}{a} arccotg \frac{bx}{a} + c = −\frac{1}{ab} arccotg \frac{bx}{a} + c, \end{matrix} for a, b∈R − \{0\}, x∈R.

∫ \frac{dx}{b^{2}x^{2} − a^{2}} = ∫ \frac{dx}{b^{2} (x^{2} − \frac{a^{2}}{b^{2}} )} = \frac{1}{b^{2}} \frac{b}{2a} ln |\frac{x − \frac{a}{b}}{x + \frac{a}{b}} | + c = \frac{1}{2ab} ln |\frac{bx − a}{bx + a}| + c, for a, b∈R − \{0\}, x∈R − \{±\frac{a}{b}\}.

∫ \frac{dx}{a^{2} − b^{2}x^{2}} = − ∫ \frac{dx}{b^{2}x^{2} − a^{2}} = −\frac{1}{2ab} ln |\frac{bx − a}{bx + a}| + c = \frac{1}{2ab} ln |\frac{bx + a}{bx − a}| + c, for a, b∈R − \{0\}, x∈R − \{±\frac{a}{b}\}.

∫ \frac{dx}{\sqrt{x^{2} + qx + r}} = \{ \begin{matrix}ln (2x + q + 2\sqrt{x^{2} + qx + r}) + c, & for q, r ∈R, 4r − q^{2} > 0, x∈R, \\ ln |2x + q| + c, & for q, r ∈R, 4r − q^{2} = 0, x∈R − \{−\frac{q}{2}\} , \\ ln |2x + q + 2\sqrt{x^{2} + qx + r}| + c, & for q, r ∈R, 4r − q^{2} < 0, x∈R, x^{2} + qx + r > 0. \end{matrix}

x^{2} + qx + r = (x^{2} + \frac{q}{2})^{2} + r − \frac{q^{2}}{4} = (x^{2} + \frac{q}{2})^{2} + \frac{4r − q^{2}}{4} , x + \frac{q}{2} + \sqrt{x^{2} + qx + r} = \frac{2x + q + 2\sqrt{x^{2} + qx + r}}{2}

∫ \frac{dx}{\sqrt{px^{2} + qx + r}} = \{ \begin{matrix}\frac{1}{\sqrt{p}} ln (2px + q + 2\sqrt{p}\sqrt{px^{2} + qx + r}) + c, & for p, q, r ∈R, p > 0, 4pr − q^{2} > 0, x∈R, \\ \frac{1}{\sqrt{p}} ln |2px + q| + c, for p, q, r ∈R, p > 0, 4pr − q^{2} = 0, x∈R − \{−\frac{q}{2p}\} , \\ \frac{1}{\sqrt{p}} ln |2px + q + 2\sqrt{p}\sqrt{px^{2} + qx + r}| + c, \\ for p, q, r ∈R, p > 0, 4pr − q^{2} < 0, x∈R, px^{2} + qx + r > 0, \\ \{ & −\frac{1}{\sqrt{−p}} arcsin \frac{2px + q}{\sqrt{q^{2} − 4pr}} + c, for p, q, r ∈R, p < 0, 4pr − q^{2} < 0, x∈R, px^{2} + qx + r > 0. \frac{1}{\sqrt{−p}} arccos \frac{2px + q}{\sqrt{q^{2} − 4pr}} + c, \end{matrix}

px^{2} + qx + r = p (x^{2} + \frac{q}{p}x + \frac{r}{p}) = p [(x + \frac{q}{2p})^{2} + \frac{r}{p} − \frac{q^{2}}{4p^{2}} ] = p (x + \frac{q}{2p})^{2} + \frac{4pr − q^{2}}{4p} = p (x + \frac{q}{2p})^{2} − \frac{q^{2} − 4pr}{4p}

p > 0: \sqrt{p} (x + \frac{q}{2p}) + \sqrt{px^{2} + qx + r} = x\sqrt{p} + \frac{q}{2\sqrt{p}} + \sqrt{px^{2} + qx + r} = \frac{2px + q + 2\sqrt{p}\sqrt{px^{2} + qx + r}}{2\sqrt{p}}

p < 0: p (x + \frac{q}{2p})^{2}− \frac{q^{2} − 4pr}{4p} = \frac{q^{2} − 4pr}{−4p} − (−p) (x + \frac{q}{2p})^{2} , \begin{matrix}\sqrt{−p} & (x + \frac{q}{2p}) \\ √ & \overline{\frac{q^{2}−4pr}{−4p}} \end{matrix} = \frac{−2p (x + \frac{q}{2p})}{\sqrt{q^{2} − 4pr}} = −\frac{2px + q}{\sqrt{q^{2} − 4pr}}

\underline{Integrals}

Substitution

Theorem.

f(x)∈C(I), x = φ(t), t∈J, φ(I) ⊂ I, φ′(t)∈C(J) ⟹ ∀t∈J: ∫ f(x) dx = ∫ f[φ(t)] dφ(t) = ∫ f[φ(t)]φ′(t) dt.

∫ \sqrt{1 − x^{2}} dx = x = sin t, t = arcsin x ⟹ dx = cos t dt \sqrt{1 − x^{2}} = \sqrt{1 − sin^{2} t} = \sqrt{cos^{2} t} = |cos t| = cos t, x∈(−1 ; 1) ⇒ t∈(−\frac{π}{2} ; \frac{π}{2} ) = ∫ cos^{2} t dt =

= ∫ \frac{1 + cos 2t}{2} dt = \frac{t}{2} + \frac{sin 2t}{4} + c = \frac{t}{2} + \frac{2 sin t cos t}{4} + c = \frac{arcsin x}{2} + \frac{x\sqrt{1 − x^{2}}}{2} + c, for x∈(−1 ; 1).

∫ sin^{3} t cos t dt = x = sin t ; dx = cos t dt = ∫ x^{3} dx = \frac{x^{4}}{4} + c = \frac{sin^{4} t}{4} + c,  for t∈R.

∫ \frac{dx}{\sqrt{x + 1} + \sqrt[3]{x + 1}} = ∫ \frac{dx}{(\sqrt[6]{x + 1})^{3} + (\sqrt[6]{x + 1})^{2}} = \begin{matrix}\sqrt[6]{x + 1} = t \\ x + 1 = t^{6} \\ dx = 6t^{5} dt \end{matrix} = ∫ \frac{6t^{5} dt}{t^{3} + t^{2}} = 6 ∫ \frac{t^{3} dt}{t + 1} =

= 6 ∫ \frac{t^{3} + t^{2} − t^{2} − t + t + 1 − 1}{t + 1} dt = 6 ∫ [\frac{t^{3} + t^{2}}{t + 1} − \frac{t^{2} + t}{t + 1} + \frac{t + 1}{t + 1} − \frac{1}{t + 1}] dt =

= 6 ∫ [t^{2} − t + 1 − \frac{1}{t + 1}] dt = 6 [\frac{t^{3}}{3} − \frac{t^{2}}{2} + t − ln |t + 1|] + c =

= 2 (\sqrt[6]{x + 1})^{3} − 3 (\sqrt[6]{x + 1})^{2} + 6\sqrt[6]{x + 1} − 6 ln |1 + \sqrt[6]{x + 1}| + c =

= 2\sqrt{x + 1} − 3\sqrt[3]{x + 1} + 6\sqrt[6]{x + 1} − 6 ln |1 + \sqrt[6]{x + 1}| + c, for x > −1.

∫ \frac{dx}{\sqrt{5 + 4 e^{x}}} = \begin{matrix}e^{x} = t, t > 0 \\ x = ln t, dx = \frac{dt}{t} \end{matrix} = ∫ \frac{dt}{t\sqrt{5 + 4t}} = \begin{matrix}4t = u^{2} \\ 4 dt = 2u du \end{matrix} = ∫ \frac{4}{u^{2}\sqrt{5 + u^{2}}} \frac{u du}{2} = 2 ∫ \frac{du}{u\sqrt{5 + u^{2}}} =

= \begin{matrix}u = \frac{\sqrt{5}}{v} ⟹ du = −\frac{\sqrt{5} dv}{v^{2}} , u > 0 \\ \sqrt{5 + u^{2}} = \sqrt{5 + \frac{5}{v^{2}}} = \frac{\sqrt{5}}{v}\sqrt{v^{2} + 1}, v > 0 \end{matrix} = 2 ∫ \frac{v}{\sqrt{5}} \frac{v}{\sqrt{5}\sqrt{v^{2} + 1}} \frac{−\sqrt{5} dv}{v^{2}} = −\frac{2}{\sqrt{5}} ∫ \frac{dv}{\sqrt{v^{2} + 1}} =

= −\frac{2}{\sqrt{5}} ln (v + \sqrt{v^{2} + 1}) + c_{1} = −\frac{2}{\sqrt{5}} ln (\frac{\sqrt{5}}{u} + \sqrt{\frac{5}{u^{2}} + 1}) + c_{1} = −\frac{2}{\sqrt{5}} ln (\frac{\sqrt{5}}{u} + \frac{\sqrt{5 + u^{2}}}{u} ) + c_{1} =

= −\frac{2}{\sqrt{5}} ln \frac{\sqrt{5} + \sqrt{5 + u^{2}}}{u} + c_{1} = \frac{2}{\sqrt{5}} [ln u − ln (\sqrt{5} + \sqrt{5 + u^{2}})] + c_{1} = \frac{2}{\sqrt{5}} [ln (4t)^{\frac{1}{2}} − ln (\sqrt{5} + \sqrt{5 + 4t})] + c_{1} =

= \frac{2}{\sqrt{5}} [\frac{1}{2} ln (4 e^{x}) − ln (\sqrt{5} + \sqrt{5 + 4 e^{x}})] + c_{1} = \frac{2}{\sqrt{5}} [\frac{1}{2} ln 4 + \frac{1}{2} ln e^{x} − ln (\sqrt{5} + \sqrt{5 + 4 e^{x}})] + c_{1} =

= \frac{ln 4}{\sqrt{5}} + \frac{x}{\sqrt{5}} − \frac{2}{\sqrt{5}} ln (\sqrt{5} + \sqrt{5 + 4 e^{x}}) + c_{1} = \frac{x}{\sqrt{5}} − \frac{2}{\sqrt{5}} ln (\sqrt{5} + \sqrt{5 + 4 e^{x}}) + c, for x∈R.

∫ \frac{dx}{\sqrt{5 + 4 e^{x}}} = e^{x} = t, 4t = u^{2}, u = \frac{\sqrt{5}}{v} ⟹ e^{x} = t = \frac{u^{2}}{4} = \frac{5}{4v^{2}} , v^{2} = \frac{5}{4 e^{x}} v = \frac{\sqrt{5}}{2\sqrt{e^{x}}} , x = ln \frac{5}{4v^{2}} = ln 5 − ln 4 − 2 ln v ⟹ dx = \frac{−2 dv}{v} = ∫ \frac{1}{\sqrt{5 + 4\frac{5}{4v^{2}}}} \frac{−2 dv}{v} =

= −\frac{2}{\sqrt{5}} ∫ \frac{dv}{\sqrt{v^{2} + 1}} = −\frac{2}{\sqrt{5}} ln (v + \sqrt{v^{2} + 1}) + c_{1} = −\frac{2}{\sqrt{5}} ln (\frac{\sqrt{5}}{2\sqrt{e^{x}}} + \sqrt{\frac{5}{4 e^{x}} + 1}) + c_{1} =

= −\frac{2}{\sqrt{5}} ln \frac{\sqrt{5} + \sqrt{5 + 4 e^{x}}}{2\sqrt{e^{x}}} + c_{1} = −\frac{2}{\sqrt{5}} ln \frac{\sqrt{5} + \sqrt{5 + 4 e^{x}}}{2 (e^{x})^{\frac{1}{2}}} + c_{1} = −\frac{2}{\sqrt{5}} [ln (\sqrt{5} + \sqrt{5 + 4 e^{x}}) − ln 2 − \frac{1}{2} ln e^{x}] + c_{1} =

= \frac{2 ln 2}{\sqrt{5}} + \frac{x}{\sqrt{5}} − \frac{2}{\sqrt{5}} ln (\sqrt{5} + \sqrt{5 + 4 e^{x}}) + c_{1} = \frac{x}{\sqrt{5}} − \frac{2}{\sqrt{5}} ln (\sqrt{5} + \sqrt{5 + 4 e^{x}}) + c, for x∈R.

∫ f (x, \sqrt[m]{\frac{ax + b}{cx + d}}) dx for ad − bc≠0

Substitution t = \sqrt[m]{\frac{ax + b}{cx + d}} ⟹ t^{m} = \frac{ax + b}{cx + d} ⟹ x = \frac{dt^{m} − b}{a − ct^{m}} ⟹ dx = \frac{mt^{m−1}(ad − bc)}{(a − ct^{m})^{2}} dt

∫ \frac{dx}{\sqrt{x + 1} + \sqrt[3]{x + 1}} = ∫ \frac{dx}{(\sqrt[6]{x + 1})^{3} + (\sqrt[6]{x + 1})^{2}} = \begin{matrix}\sqrt[6]{x + 1} = t \\ x + 1 = t^{6} \\ dx = 6t^{5} dt \end{matrix} = ∫ \frac{6t^{5} dt}{t^{3} + t^{2}} = 6 ∫ \frac{t^{3} dt}{t + 1} =

= 6 ∫ \frac{t^{3} + t^{2} − t^{2} − t + t + 1 − 1}{t + 1} dt = 6 ∫ [\frac{t^{3} + t^{2}}{t + 1} − \frac{t^{2} + t}{t + 1} + \frac{t + 1}{t + 1} − \frac{1}{t + 1}] dt = 6 ∫ [t^{2} − t + 1 − \frac{1}{t + 1}] dt =

= 6 [\frac{t^{3}}{3} − \frac{t^{2}}{2} + t − ln |t + 1|] + c = 2 (\sqrt[6]{x + 1})^{3} − 3 (\sqrt[6]{x + 1})^{2} + 6\sqrt[6]{x + 1} − 6 ln |1 + \sqrt[6]{x + 1}| + c =

= 2\sqrt{x + 1} − 3\sqrt[3]{x + 1} + 6\sqrt[6]{x + 1} − 6 ln |1 + \sqrt[6]{x + 1}| + c, for x > −1.

∫ \begin{matrix}√ & \overline{1 − \sqrt{x}} \\ √ & \overline{1 + \sqrt{x}} \end{matrix} dx = ∫ \begin{matrix}√ & \overline{1 − \sqrt{x}} \\ √ & \overline{1 + \sqrt{x}} \end{matrix} · \begin{matrix}√ & \overline{1 − \sqrt{x}} \\ √ & \overline{1 − \sqrt{x}} \end{matrix} dx = ∫ \frac{1 − \sqrt{x}}{\sqrt{1 − x}} dx = ∫ \frac{dx}{\sqrt{1 − x}} − ∫ \begin{matrix}√ & \overline{x} \\ √ & \overline{1 − x} \end{matrix} dx =

\begin{matrix}1 − x = u \\ dx = − du \end{matrix} resp.  \frac{x}{1 − x} = t^{2} ⟹ x = t^{2} − xt^{2} ⟹ x = \frac{t^{2}}{1 + t^{2}} ⟹ dx = \frac{2t(1 + t^{2}) − t^{2} · 2t}{(1 + t^{2})^{2}} dt = \frac{2t dt}{(1 + t^{2})^{2}}

= ∫ (1 − x)^{− \frac{1}{2}} dx − ∫ \sqrt{\frac{x}{1 − x}} dx = − ∫ u^{− \frac{1}{2}} du − ∫ \frac{t · 2t dt}{(1 + t^{2})^{2}} = −\frac{u^{\frac{1}{2}}}{\frac{1}{2}} − 2 ∫ \frac{t^{2} dt}{(1 + t^{2})^{2}} = −2\sqrt{u} − 2 ∫ \frac{1 + t^{2} − 1}{(1 + t^{2})^{2}} dt =

= −2\sqrt{1 − x} − 2 ∫ [\frac{1}{1 + t^{2}} − \frac{1}{(1 + t^{2})^{2}} ] dt = −2\sqrt{1 − x} − 2 [arctg t − \frac{1}{2} arctg t − \frac{1}{2} \frac{t}{t^{2} + 1}] + c =

t^{2} + 1 = \frac{x}{1 − x} + 1 = \frac{x + 1 − x}{1 − x} = \frac{1}{1 − x} ⟹ \frac{t}{t^{2} + 1} = (1 − x)\sqrt{\frac{x}{1 − x}} = \sqrt{(1 − x)x} = \sqrt{x − x^{2}}

= −2\sqrt{1 − x} − arctg t + \frac{t}{t^{2} + 1} + c = −2\sqrt{1 − x} − arctg \sqrt{\frac{x}{1 − x}} + \sqrt{x − x^{2}} + c, for x∈⟨0 ; 1).

∫ f(x, \sqrt{ax^{2} + bx + c}) dx for a≠0

First Euler substitution: \sqrt{ax^{2} + bx + c} = t ± x\sqrt{a} for a > 0 ⟹ t = \sqrt{ax^{2} + bx + c} ∓ x\sqrt{a}

⟹ ax^{2} + bx + c = t^{2} ± 2tx\sqrt{a} + ax^{2} ⟹ x = \frac{t^{2} − c}{b ∓ 2t\sqrt{a}} ⟹ dx = 2\frac{∓t^{2}\sqrt{a} + tb ∓ c\sqrt{a}}{(b ∓ 2t\sqrt{a})^{2}} dt

Second Euler substitution: \sqrt{ax^{2} + bx + c} = xt ± \sqrt{c} for c > 0 ⟹ t = \frac{\sqrt{ax^{2} + bx + c} ∓ \sqrt{c}}{x}

⟹ ax^{2} + bx + c = x^{2}t^{2} ± 2tx\sqrt{c} + c ⟹^{x≠0} x = \frac{±2t\sqrt{c} − b}{a − t^{2}} ⟹ dx = 2\frac{±t^{2}\sqrt{c} − tb ± a\sqrt{c}}{(a − t^{2})^{2}} dt

Third Euler substitution: t = \sqrt{a\frac{x − u}{x − v}}, if u, v are real roots of the ax^{2} + bx + c = 0 ⟹ x = \frac{vt^{2} − au}{t^{2} − a} ⟹ dx = \frac{2ta(u − v)}{(t^{2} − a)^{2}} dt

∫ \frac{dx}{\sqrt{x^{2} + a^{2}}} = 1-st:  \sqrt{x^{2} + a^{2}} = t − x ⟹ x^{2} + a^{2} = t^{2} − 2tx + x^{2} ⟹ 2tx = t^{2} − a^{2} ⟹ x = \frac{t^{2} − a^{2}}{2t} \sqrt{x^{2} + a^{2}} = t − \frac{t^{2} − a^{2}}{2t} = \frac{t^{2} + a^{2}}{2t} ,  dx = \frac{2t · 2t − 2(t^{2} − a^{2})}{4t^{2}} dt = \frac{2t^{2} + 2a^{2}}{4t^{2}} dt = \frac{t^{2} + a^{2}}{2t^{2}} dt \sqrt{a^{2} + x^{2}} > \sqrt{x^{2}} ≥ x ⟹ \sqrt{a^{2} + x^{2}} − x > 0 ⟹ ln |x + \sqrt{a^{2} + x^{2}}| = ln (x + \sqrt{a^{2} + x^{2}}) =

= ∫ \frac{2t}{t^{2} + a^{2}} \frac{t^{2} + a^{2}}{2t^{2}} dt = ∫ \frac{dt}{t} = ln |t| + c = ln |x + \sqrt{x^{2} + a^{2}}| + c = ln (x + \sqrt{x^{2} + a^{2}}) + c, for a∈R, x∈R.

∫ \frac{dx}{\sqrt{a^{2} − x^{2}}} = \begin{matrix}2-nd:  \sqrt{a^{2} − x^{2}} = xt + a ⟹ a^{2} − x^{2} = x^{2}t^{2} + 2txa + a^{2} ⟹^{x≠0} x = \frac{−2ta}{t^{2} + 1}, t = \frac{\sqrt{a^{2} − x^{2}} − a}{x} \\ \sqrt{a^{2} − x^{2}} = \frac{−2t^{2}a}{t^{2} + 1} + a = \frac{a(1 − t^{2})}{t^{2} + 1},  dx = \frac{−2a(t^{2} + 1) + 2ta · 2t}{(t^{2} + 1)^{2}} dt = \frac{2a(t^{2} − 1)}{(t^{2} + 1)^{2}} dt \end{matrix} =

= ∫ \frac{t^{2} + 1}{a(1 − t^{2})} \frac{2a(t^{2} − 1)}{(t^{2} + 1)^{2}} dt = −2 ∫ \frac{dt}{t^{2} + 1} = −2 arctg t + c = −2 arctg \frac{\sqrt{a^{2} − x^{2}} − a}{x} + c,

for a∈R, a≠0, x∈(− |a|; |a|) − {0}.

∫ \frac{dx}{\sqrt{x(1 − x)}} = \begin{matrix}3-rd: t = \sqrt{\frac{x}{1 − x}} > 0 ⟹ t^{2} = \frac{x}{1 − x} ⟹ t^{2} − t^{2}x = x ⟹ x = \frac{t^{2}}{t^{2} + 1} \\ dx = \frac{2t(t^{2}+1) − t^{2}·2t}{(t^{2} + 1)^{2}} dt = \frac{2t dt}{(t^{2}+1)^{2}} , x(1−x) = \frac{t^{2}}{t^{2}+1} (1− \frac{t^{2}}{t^{2}+1}) = \frac{t^{2}}{(t^{2}+1)^{2}} \end{matrix} = ∫ \frac{t^{2} + 1}{t} \frac{2t dt}{(t^{2} + 1)^{2}} =

= ∫ \frac{2 dt}{t^{2} + 1} = 2 arctg t + c = 2 arctg \sqrt{\frac{x}{1 − x}} + c, for x(1 − x) = x − x^{2} > 0 ⇒ x > x^{2} ⇒ x∈(0 ; 1).

∫ f(sin x, cos x) dx

Universal trig substitution: t = tg \frac{x}{2} ⟹ x = 2 arctg t ⟹ dx = \frac{2 dt}{t^{2} + 1}

⟹ sin x = 2 sin \frac{x}{2} cos \frac{x}{2} = \frac{2 sin \frac{x}{2} cos \frac{x}{2}}{sin^{2} \frac{x}{2} + cos^{2} \frac{x}{2}} = \frac{2 sin \frac{x}{2} cos \frac{x}{2}}{sin^{2} \frac{x}{2} + cos^{2} \frac{x}{2}} · \begin{matrix}cos & −2 & \frac{x}{2} \\ cos & −2 & \frac{x}{2} \end{matrix} = \frac{2 tg \frac{x}{2}}{tg^{2} \frac{x}{2} + 1} = \frac{2t}{t^{2} + 1}

⟹ cos x = cos^{2} \frac{x}{2} − sin^{2} \frac{x}{2} = \frac{cos^{2} \frac{x}{2} − sin^{2} \frac{x}{2}}{sin^{2} \frac{x}{2} + cos^{2} \frac{x}{2}} = \frac{cos^{2} \frac{x}{2} − sin^{2} \frac{x}{2}}{sin^{2} \frac{x}{2} + cos^{2} \frac{x}{2}} · \begin{matrix}cos & −2 & \frac{x}{2} \\ cos & −2 & \frac{x}{2} \end{matrix} = \frac{1 − tg^{2} \frac{x}{2}}{tg^{2} \frac{x}{2} + 1} = \frac{1 − t^{2}}{t^{2} + 1}

∫ \frac{dx}{cos x} = t = tg \frac{x}{2}, dx = \frac{2 dt}{1 + t^{2}} , cos x = \frac{1 − t^{2}}{1 + t^{2}} = ∫ \frac{1 + t^{2}}{1 − t^{2}} \frac{2 dt}{1 + t^{2}} = ∫ \frac{2 dt}{1 − t^{2}} = − ∫ \frac{2 dt}{t^{2} − 1} = −\frac{2}{2} ln |\frac{t − 1}{t + 1}| + c =

= ln |\frac{t + 1}{t − 1}| + c = ln |\frac{tg \frac{x}{2} + 1}{tg \frac{x}{2} − 1}| + c, for x∈R − \{(2k + 1)\frac{π}{2} ; k ∈Z\}.

∫ \frac{dx}{1 + cos x} = t = tg \frac{x}{2}, dx = \frac{2 dt}{1 + t^{2}} , 1 + cos x = 1 + \frac{1 − t^{2}}{1 + t^{2}} = \frac{2}{1 + t^{2}} = ∫ \frac{1 + t^{2}}{2} \frac{2 dt}{1 + t^{2}} = ∫ dt = t + c = tg \frac{x}{2} + c,

for x∈R − {π + 2kπ; k ∈Z}.

∫ \frac{dx}{sin x} = t = tg \frac{x}{2}, dx = \frac{2 dt}{1 + t^{2}} , sin x = \frac{2t}{1 + t^{2}} = ∫ \frac{1 + t^{2}}{2t} \frac{2 dt}{1 + t^{2}} = ∫ \frac{dt}{t} = ln |t|+c = ln |tg \frac{x}{2}| + c, for x∈R−\{kπ; k ∈Z\}.

∫ \frac{dx}{1 + sin x} = t = tg \frac{x}{2}, dx = \frac{2 dt}{1 + t^{2}} , 1 + sin x = 1 + \frac{2t}{1 + t^{2}} = \frac{1 + 2t + t^{2}}{1 + t^{2}} = ∫ \frac{1 + t^{2}}{(1 + t)^{2}} \frac{2 dt}{1 + t^{2}} = ∫ \frac{2 dt}{(1 + t)^{2}} =

= \begin{matrix}1 + t = u \\ dt = du \end{matrix} = 2 ∫ \frac{du}{u^{2}} = 2 ∫ u^{−2} du = 2\frac{u^{−1}}{−1} + c = c − \frac{2}{t + 1} = c − \frac{2}{tg \frac{x}{2} + 1}, for x∈R − \{\frac{π}{2} + kπ; k ∈Z\}.

\underline{Integrals}

Integration by parts (per partes)

Theorem.

Let f, g be function differentiable on an interval I. Then fg′ is integrable on I if and only if f′g is integrable on I, and ∀x∈I: ∫ f′(x)g(x) dx = f(x)g(x) − ∫ f(x)g′(x) dx.

∫ arctg x dx = \begin{matrix}u′ = 1 ⇒ & u = x \\ v = arctg x ⇒ & v′ = \frac{1}{1 + x^{2}} \end{matrix} = x arctg x − ∫ \frac{x}{1 + x^{2}} dx = x arctg x − \frac{1}{2} ∫ \frac{2x}{1 + x^{2}} dx =

= x arctg x − \frac{1}{2} ln |1 + x^{2}| + c = x arctg x − \frac{1}{2} ln (1 + x^{2}) + c = x arctg x − ln \sqrt{1 + x^{2}} + c, for x∈R.

∫ x^{2} ln x dx = \begin{matrix}u = ln x ⇒ & u′ = \frac{1}{x} \\ v′ = x^{2} ⇒ & v = \frac{x^{3}}{3} \end{matrix} = \frac{x^{3}}{3} ln x −∫ \frac{x^{3}}{3} \frac{1}{x} dx = \frac{x^{3}}{3} ln x −∫ \frac{x^{2}}{3} dx = \frac{x^{3}}{3} ln x − \frac{x^{3}}{9} + c, for x > 0.

∫ \frac{ln x}{\sqrt{x}} dx = \begin{matrix}u = ln x ⇒ & u′ = \frac{1}{x} \\ v′ = x^{− \frac{1}{2}} ⇒ & v = 2x^{\frac{1}{2}} \end{matrix} = 2\sqrt{x} ln x − 2 ∫ \frac{x^{\frac{1}{2}}}{x} dx = 2\sqrt{x} ln x − 2 ∫ x^{− \frac{1}{2}} dx =

= 2\sqrt{x} ln x − 2 · 2x^{\frac{1}{2}} + c = 2\sqrt{x} ln x − 4\sqrt{x} + c, for x > 0.

∫ x ln^{2} x dx = \begin{matrix}u = ln^{2} x ⇒ & u′ = \frac{2 ln x}{x} \\ v′ = x ⇒ & v = \frac{x^{2}}{2} \end{matrix} = \frac{x^{2}}{2} ln^{2} x − ∫ \frac{x^{2}}{2} \frac{2 ln x}{x} dx = \frac{x^{2}}{2} ln^{2} x − ∫ x ln x dx = \begin{matrix}u = ln x ⇒ & u′ = \frac{1}{x} \\ v′ = x ⇒ & v = \frac{x^{2}}{2} \end{matrix} =

= \frac{x^{2}}{2} ln^{2} x − [\frac{x^{2}}{2} ln x − ∫ \frac{x^{2}}{2} \frac{1}{x} dx] = \frac{x^{2}}{2} ln^{2} x − \frac{x^{2}}{2} ln x + ∫ \frac{x dx}{2} = \frac{x^{2}}{2} ln^{2} x − \frac{x^{2}}{2} ln x + \frac{x^{2}}{4} + c, for x > 0.

∫ sin^{n} x dx = ∫ sin x sin^{n−1} x dx = u = sin^{n−1} x ⇒ u′ = (n − 1) sin^{n−2} x cos x v′ = sin x ⇒ v = − cos x =

= − cos x sin^{n−1} x + (n − 1) ∫ sin^{n−2} x cos^{2} x dx = − cos x sin^{n−1} x + (n − 1) ∫ sin^{n−2} x(1 − sin^{2} x) dx =

= − cos x sin^{n−1} x + (n − 1) ∫ sin^{n−2} x dx − (n − 1) ∫ sin^{n} x dx

⟹^{(i.e.equation)} ∫ sin^{n} x dx + (n − 1) ∫ sin^{n} x dx = − cos x sin^{n−1} x + (n − 1) ∫ sin^{n−2} x dx

⟹ ∫ sin^{n} x dx = −\frac{cos x sin^{n−1} x}{n} + \frac{n − 1}{n} ∫ sin^{n−2} x dx, for n = 3, 4, 5, …, x∈R.

∫ cos^{n} x dx = ∫ cos x cos^{n−1} x dx = u = cos^{n−1} x ⇒ u′ = −(n − 1) cos^{n−2} x sin x v′ = cos x ⇒ v = sin x =

= sin x cos^{n−1} x + (n − 1) ∫ cos^{n−2} x sin^{2} x dx = sin x cos^{n−1} x + (n − 1) ∫ cos^{n−2} x(1 − cos^{2} x) dx =

= sin x cos^{n−1} x + (n − 1) ∫ cos^{n−2} x dx − (n − 1) ∫ cos^{n} x dx

⟹^{(i.e.equation)} ∫ cos^{n} x dx + (n − 1) ∫ cos^{n} x dx = sin x cos^{n−1} x + (n − 1) ∫ cos^{n−2} x dx

⟹ ∫ cos^{n} x dx = \frac{sin x cos^{n−1} x}{n} + \frac{n − 1}{n} ∫ cos^{n−2} x dx, for n = 3, 4, 5, …, x∈R.

\underline{Integrals}

Partial fractions

∫ \frac{dx}{(x − a)^{n}} = \begin{matrix}x − a = t \\ dx = du \end{matrix} = ∫ \frac{dt}{t^{n}} = \{ \begin{matrix}∫ \frac{dt}{t} = ln |t| + c = ln |x − a| + c, & for n = 1, \\ ∫ t^{−n} dt = \frac{t^{1−n}}{1 − n} + c = \frac{(x − a)^{1−n}}{1 − n} + c, & for n = 2, 3, 4, …, \end{matrix} for a∈R, x∈R − \{a\}.

∫ \frac{dx}{x + 1} = ln |x + 1| + c, for x∈R − \{−1\}.

∫ \frac{1}{(x − 1)^{n}} dx = \begin{matrix}x − 1 = t \\ dx = dt \end{matrix} = ∫ t^{−n} dt = \frac{t^{1−n}}{1 − n} + c = \frac{(x − 1)^{1−n}}{1 − n} + c, for n = 2, 3, 4, …, x∈R − \{1\}.

∫ \frac{dx}{x^{2} + ax + b} = \begin{matrix}x + \frac{a}{2} = t, α^{2} = b − \frac{a^{2}}{4} > 0 \\ dx = dt, α = \sqrt{b − a^{2}/4} \end{matrix} = ∫ \frac{dt}{t^{2} + α^{2}} = \frac{1}{α} arctg \frac{t}{α} + c = \frac{1}{\sqrt{b − \frac{a^{2}}{4}}} arctg \frac{x + \frac{a}{2}}{\sqrt{b − \frac{a^{2}}{4}}} + c.

for a, b, A∈R, a^{2} − 4b < 0, x∈R.

∫ \frac{dx}{x^{2} + 4x + 25} = \begin{matrix}x^{2} + 4x + 25 = (x + 2)^{2} + 21 \\ x + 2 = t, dx = dt \end{matrix} = \frac{1}{\sqrt{25 − \frac{4^{2}}{4}}} arctg \frac{x + \frac{4}{2}}{\sqrt{25 − \frac{4^{2}}{4}}} + c = \frac{1}{\sqrt{21}} arctg \frac{x + 2}{\sqrt{21}} + c, for x∈R.

∫ \frac{dx}{x^{2} + b} = \frac{1}{\sqrt{b − \frac{0^{2}}{4}}} arctg \frac{x + \frac{0}{2}}{\sqrt{b − \frac{0^{2}}{4}}} + c = \frac{1}{\sqrt{b}} arctg \frac{x}{\sqrt{b}} + c, for b > 0, x∈R.

∫ \frac{dx}{x^{2} + β^{2}} = \frac{1}{|β|} arctg \frac{x}{|β|} + c = \frac{1}{−β} arctg \frac{x}{−β} = −\frac{1}{β} [− arctg \frac{x}{β} ] = \frac{1}{β} arctg \frac{x}{β} = \frac{1}{β} arctg \frac{x}{β} + c, for β ∈R, x∈R.

∫ \frac{dx}{x^{2} + 1} = arctg x + c, for x∈R.

∫ \frac{x + A}{(x^{2} + ax + b)^{n}} dx = \frac{1}{2} ∫ \frac{2x + a + 2A − a}{(x^{2} + ax + b)^{n}} dx = \frac{1}{2} ∫ \frac{2x + a}{(x^{2} + ax + b)^{n}} dx + \frac{2A − a}{2} ∫ \frac{dx}{(x^{2} + ax + b)^{n}} =

= \{ \begin{matrix}\frac{1}{2} ln |x^{2} + ax + b| + \frac{2A − a}{2\sqrt{b − \frac{a^{2}}{4}}} arctg \frac{x + \frac{a}{2}}{\sqrt{b − \frac{a^{2}}{4}}} + c, & for n = 1, \\ \frac{1}{2(1 − n)(x^{2} + ax + b)^{n−1}} − \frac{2A − a}{4 (b − \frac{a^{2}}{4} ) (1 − n)} \frac{x + \frac{a}{2}}{(x^{2} + ax + b)^{n−1}} + \frac{(3 − 2n)(2A − a)}{4 (b − \frac{a^{2}}{4} ) (1 − n)} ∫ \frac{dx}{(x^{2} + ax + b)^{n−1}} , & for n = 2, 3, 4, …, \end{matrix}

for a, b, A∈R, a^{2} − 4b < 0 (i.e. x^{2} + ax + b = 0 has no real roots), x∈R.

∫ \frac{2x + a}{(x^{2} + ax + b)^{n}} dx = \begin{matrix}x^{2} + ax + b = u \\ (2x + a) dx = du \end{matrix} = \{ \begin{matrix}∫ \frac{du}{u} = ln |u| + c = ln |x^{2} + ax + b| + c, & for n = 1, \\ ∫ \frac{du}{u^{n}} = ∫ u^{−n} du = \frac{u^{1−n}}{1 − n} + c = \frac{1}{(1 − n)(x^{2} + ax + b)^{n−1}} + c, & for n = 2, 3, 4, …. \end{matrix}

for a, b∈R, x∈R, x≠ \frac{−a ± \sqrt{a^{2} − 4b}}{2}.

∫ \frac{dx}{(x^{2} + ax + b)^{n}} = \begin{matrix}x^{2} + ax + b = (x + \frac{a}{2})^{2} + b − \frac{a^{2}}{4} , x + \frac{a}{2} = t \\ α^{2} = b − \frac{a^{2}}{4} > 0, α = \sqrt{b − a^{2}/4}, dx = dt \end{matrix} = ∫ \frac{dt}{(t^{2} + α^{2})^{n}} = \frac{1}{α^{2}} ∫ \frac{α^{2}}{(t^{2} + α^{2})^{n}} dt =

= \frac{1}{α^{2}} ∫ \frac{t^{2} + α^{2} − t^{2}}{(t^{2} + α^{2})^{n}} dt = \frac{1}{α^{2}} ∫ \frac{t^{2} + α^{2}}{(t^{2} + α^{2})^{n}} dt − \frac{1}{α^{2}} ∫ \frac{t^{2}}{(t^{2} + α^{2})^{n}} dt = \frac{1}{α^{2}} ∫ \frac{dt}{(t^{2} + α^{2})^{n−1}} − \frac{1}{α^{2}} ∫ \frac{t · t}{(t^{2} + α^{2})^{n}} dt =

u = t, u′ = 1 ⟹ v′ = \frac{t}{(t^{2} + α^{2})^{n}} = t(t^{2} + α^{2})^{−n}, v = \frac{(t^{2} + α^{2})^{1−n}}{2(1 − n)} = \frac{1}{2(1 − n)} \frac{1}{(t^{2} + α^{2})^{n−1}}

= \frac{1}{α^{2}} ∫ \frac{dt}{(t^{2} + α^{2})^{n−1}} − \frac{1}{α^{2}} [\frac{1}{2(1 − n)} \frac{t}{(t^{2} + α^{2})^{n−1}} − \frac{1}{2(1 − n)} ∫ \frac{dt}{(t^{2} + α^{2})^{n−1}} ] =

= \frac{1}{α^{2}} [1 + \frac{1}{2(1 − n)}] ∫ \frac{dt}{(t^{2} + α^{2})^{n−1}} − \frac{1}{2α^{2}(1 − n)} \frac{t}{(t^{2} + α^{2})^{n−1}} =

= \frac{3 − 2n}{2α^{2}(1 − n)} ∫ \frac{dt}{(t^{2} + α^{2})^{n−1}} − \frac{1}{2α^{2}(1 − n)} \frac{t}{(t^{2} + α^{2})^{n−1}} =

= \frac{3 − 2n}{2 (b − \frac{a^{2}}{4} ) (1 − n)} ∫ \frac{dx}{(x^{2} + ax + b)^{n−1}} − \frac{1}{2 (b − \frac{a^{2}}{4} ) (1 − n)} \frac{x + \frac{a}{2}}{(x^{2} + ax + b)^{n−1}} ,

for a, b∈R, a^{2} − 4b < 0, n = 2, 3, 4, …, x∈R.

∫ \frac{dx}{(x^{2} + ax + b)^{2}} = \frac{1}{2 (b − \frac{a^{2}}{4} )} ∫ \frac{dx}{x^{2} + ax + b} + \frac{1}{2 (b − \frac{a^{2}}{4} )} \frac{x + \frac{a}{2}}{x^{2} + ax + b} = \frac{1}{2 (b − \frac{a^{2}}{4} )^{\frac{3}{2}}} arctg \frac{x + \frac{a}{2}}{\sqrt{b − \frac{a^{2}}{4}}} + \frac{1}{2 (b − \frac{a^{2}}{4} )} \frac{x + \frac{a}{2}}{x^{2} + ax + b} + c,

for a, b∈R, a^{2} − 4b < 0, x∈R.

∫ \frac{dx}{(x^{2} + 4x + 25)^{2}} = \frac{a}{2} = \frac{4}{2} = 2, b − \frac{a^{2}}{4} = 25 − \frac{4^{2}}{4} = 25 − 4 = 21 = \frac{1}{2 · 21 · \sqrt{21}} arctg \frac{x + 2}{\sqrt{21}} + \frac{1}{2 · 21} \frac{x + 2}{x^{2} + 4x + 25} + c, for x∈R.

∫ \frac{dx}{(x^{2} + b)^{2}} = \frac{1}{2b} ∫ \frac{dx}{x^{2} + b} + \frac{1}{2b} \frac{x}{x^{2} + b} = \frac{1}{2b\sqrt{b}} arctg \frac{x}{\sqrt{b}} + \frac{1}{2b} \frac{x}{x^{2} + b} + c, for b > 0, x∈R.

∫ \frac{dx}{(x^{2} + β^{2})^{2}} = \frac{1}{2β^{3}} arctg \frac{x}{β} + \frac{1}{2β^{2}} \frac{x}{x^{2} + β^{2}} + c, for β ∈R, x∈R.

∫ \frac{dx}{(x^{2} + 1)^{2}} = \frac{1}{2} arctg x + \frac{1}{2} \frac{x}{x^{2} + 1} + c, for x∈R.

∫ \frac{dx}{(x^{2} + ax + b)^{3}} = \frac{3}{4 (b − \frac{a^{2}}{4} )} ∫ \frac{dx}{(x^{2} + ax + b)^{2}} + \frac{1}{4 (b − \frac{a^{2}}{4} )} \frac{x + \frac{a}{2}}{(x^{2} + ax + b)^{2}} .

= _{\frac{3}{4(b− \frac{a^{2}}{4} )}} [_{\frac{1}{2(b− \frac{a^{2}}{4} )^{\frac{3}{2}}}} arctg \frac{x+ \frac{a}{2}}{\sqrt{b− \frac{a^{2}}{4}}} + _{\frac{1}{2(b− \frac{a^{2}}{4} )}} \frac{x+ \frac{a}{2}}{x^{2}+ax+b}] + _{\frac{1}{4(b− \frac{a^{2}}{4} )}} \frac{x+ \frac{a}{2}}{(x^{2}+ax+b)^{2}} + c =

= _{\frac{3}{8(b− \frac{a^{2}}{4} )^{\frac{5}{2}}}} arctg \frac{x+ \frac{a}{2}}{\sqrt{b− \frac{a^{2}}{4}}} + _{\frac{3}{8(b− \frac{a^{2}}{4} )^{2}}} \frac{x+ \frac{a}{2}}{x^{2}+ax+b} + _{\frac{1}{4(b− \frac{a^{2}}{4} )}} \frac{x+ \frac{a}{2}}{(x^{2}+ax+b)^{2}} + c,

for a, b∈R, a^{2} − 4b < 0, x∈R.

∫ \frac{dx}{(x^{2} + 4x + 25)^{3}} = \frac{3}{8 · 21^{2} · \sqrt{21}} arctg \frac{x + 2}{\sqrt{21}} + \frac{3}{8 · 21^{2}} \frac{x + 2}{x^{2} + 4x + 25} + \frac{1}{4 · 21} \frac{x + 2}{(x^{2} + 4x + 25)^{2}} + c, for x∈R.

∫ \frac{dx}{(x^{2} + b)^{3}} = \frac{3}{8b^{2}\sqrt{b}} arctg \frac{x}{\sqrt{b}} + \frac{3}{8b^{2}} \frac{x}{x^{2} + b} + \frac{1}{4b} \frac{x}{(x^{2} + b)^{2}} + c, for b > 0, x∈R.

∫ \frac{dx}{(x^{2} + β^{2})^{3}} = \frac{3}{8β^{5}} arctg \frac{x}{β} + \frac{3}{8β^{4}} \frac{x}{x^{2} + β^{2}} + \frac{1}{4β^{2}} \frac{x}{(x^{2} + β^{2})^{2}} + c, for β ∈R, x∈R.

∫ \frac{dx}{(x^{2} + 1)^{3}} = \frac{3}{8} arctg x + \frac{3}{8} \frac{x}{x^{2} + 1} + \frac{1}{4} \frac{x}{(x^{2} + 1)^{2}} + c, for x∈R.

∫ \frac{dx}{(x^{2} + ax + b)^{n}} = \frac{3 − 2n}{2 (b − \frac{a^{2}}{4} ) (1 − n)} ∫ \frac{dx}{(x^{2} + ax + b)^{n−1}} − \frac{1}{2 (b − \frac{a^{2}}{4} ) (1 − n)} \frac{x + \frac{a}{2}}{(x^{2} + ax + b)^{n−1}} =

= \frac{3 − 2n}{2 (b − \frac{a^{2}}{4} ) (1 − n)} [\frac{3 − 2(n − 1)}{2 (b − \frac{a^{2}}{4} ) (1 − (n − 1))} ∫ \frac{dx}{(x^{2} + ax + b)^{(n−1)−1}} − \frac{1}{2 (b − \frac{a^{2}}{4} ) (1 − (n − 1))} \frac{x + \frac{a}{2}}{(x^{2} + ax + b)^{(n−1)−1}} ] −

−\frac{1}{2 (b − \frac{a^{2}}{4} ) (1 − n)} \frac{x + \frac{a}{2}}{(x^{2} + ax + b)^{n−1}} =

= \frac{(3 − 2n)(5 − 2n)}{2^{2} (b − \frac{a^{2}}{4} )^{2} (1 − n)(2 − n)} ∫ \frac{dx}{(x^{2} + ax + b)^{n−2}} − \frac{3 − 2n}{2^{2} (b − \frac{a^{2}}{4} )^{2} (1 − n)(2 − n)} \frac{x + \frac{a}{2}}{(x^{2} + ax + b)^{n−2}} −

−\frac{1}{2 (b − \frac{a^{2}}{4} ) (1 − n)} \frac{x + \frac{a}{2}}{(x^{2} + ax + b)^{n−1}} = ⋯,

for a, b∈R, a^{2} − 4b < 0, n = 3, 4, 5, …, x∈R.

∫ \frac{dx}{(x^{2} + b)^{n}} = \frac{3 − 2n}{2b(1 − n)} ∫ \frac{dx}{(x^{2} + b)^{n−1}} − \frac{1}{2b(1 − n)} \frac{x}{(x^{2} + b)^{n−1}} =

= \frac{3 − 2n}{2b(1 − n)} [\frac{5 − 2n}{2b(2 − n)} ∫ \frac{dx}{(x^{2} + b)^{n−2}} − \frac{1}{2b(2 − n)} \frac{x}{(x^{2} + b)^{n−2}} ] − \frac{1}{2b(1 − n)} \frac{x}{(x^{2} + b)^{n−1}} =

= \frac{(3 − 2n)(5 − 2n)}{2^{2}b^{2}(1 − n)(2 − n)} ∫ \frac{dx}{(x^{2} + b)^{n−2}} − \frac{3 − 2n}{2^{2}b^{2}(1 − n)(2 − n)} \frac{x}{(x^{2} + b)^{n−2}} − \frac{1}{2b(1 − n)} \frac{x}{(x^{2} + b)^{n−1}} = ⋯,

for b > 0, n = 3, 4, 5, …, x∈R.

∫ \frac{dx}{(x^{2} + β^{2})^{n}} = \frac{3 − 2n}{2β^{2}(1 − n)} ∫ \frac{dx}{(x^{2} + β^{2})^{n−1}} − \frac{1}{2β^{2}(1 − n)} \frac{x}{(x^{2} + β^{2})^{n−1}} , for β ∈R, n = 2, 3, 4, …, x∈R.

∫ \frac{dx}{(x^{2} + 1)^{n}} = \frac{3 − 2n}{2(1 − n)} ∫ \frac{dx}{(x^{2} + 1)^{n−1}} − \frac{1}{2(1 − n)} \frac{x}{(x^{2} + 1)^{n−1}} , for n = 2, 3, 4, …, x∈R.

∫ \frac{x dx}{(x^{2} + 1)^{n}} = \begin{matrix}x^{2} + 1 = t \\ 2x dx = dt \end{matrix} = ∫ \frac{t^{−n} dt}{2} = \{ \begin{matrix}\frac{1}{2} |t| +c = \frac{1}{2} ln |x^{2} + 1| + c = \frac{1}{2} ln (x^{2} + 1) + c = ln \sqrt{x^{2} + 1} + c, for n = 1, \\ \frac{1}{2} \frac{t^{−n+1}}{−n + 1} + c = \frac{1}{2(1 − n)t^{n−1}} + c = \frac{1}{2(1 − n) (x^{2} + 1)^{n−1}} + c, for n = 2, 3, 4, …, \end{matrix} for x∈R.

\underline{Integrals}

Fourier Series

∫_{a}^{a} f(x) dx = 0, ∫_{a}^{b} f(x) = − ∫_{b}^{a} f(x), for all a, b∈R, f(x)∈R_{⟨a; b⟩} [i.e. f(x) is Riemann integrable on ⟨a; b⟩]

Theorem.

The Fourier series of a function f(x)∈R_{⟨a; a+2π⟩}, a∈R is given by

f(x) ≈ A_{0} \frac{1}{\sqrt{2π}} + ∑_{n=1}^{∞} [A_{n} \begin{matrix}cos & nx \\ √ & \overline{π} \end{matrix} + B_{n} \begin{matrix}sin & nx \\ √ & \overline{π} \end{matrix}], for x∈⟨a; a + 2π⟩,

where A_{0} = ∫_{a}^{a+2π} \frac{f(x)}{\sqrt{2π}} dx, A_{n} = ∫_{a}^{a+2π} f(x)\begin{matrix}cos & nx \\ √ & \overline{π} \end{matrix} dx, B_{n} = ∫_{a}^{a+2π} f(x)\begin{matrix}sin & nx \\ √ & \overline{π} \end{matrix} dx, n∈N,

resp. f(x) ≈ \frac{a_{0}}{2} + ∑_{n=1}^{∞} [a_{n} cos nx + b_{n} sin nx], for x∈⟨a; a + 2π⟩,

where a_{0} = \frac{1}{π} ∫_{a}^{a+2π} f(x) dx, a_{n} = \frac{1}{π} ∫_{a}^{a+2π} f(x) cos nx dx, b_{n} = \frac{1}{π} ∫_{a}^{a+2π} f(x) sin nx dx, n∈N.

A Fourier series converges to the function f̃(x_{0}) = \{ \begin{matrix}\frac{1}{2} [ lim_{x→x_{0}^{−}} f(x) + lim_{x→x_{0}^{+}} f(x)] , & x_{0} ∈(a; a + 2π) , \\ f(x_{0} − 2kπ), & x_{0} ∈(a + 2kπ; a + 2π + 2kπ) , \\ \frac{1}{2} [ lim_{x→a^{+}} f(x) + lim_{x→a+2π^{−}} f(x)] , & x_{0} = a + 2kπ, k ∈Z. \end{matrix}

(Function f̃(x) is a periodic function with period 2π and equals to the original function at the interval ⟨a; a + 2π⟩ at points of continuity or to the average of the two limits at points of discontinuity.)

f(x) ≈ f̃(x) = A_{0} \frac{1}{\sqrt{2π}} + ∑_{n=1}^{∞} [A_{n} \begin{matrix}cos & nx \\ √ & \overline{π} \end{matrix} + B_{n} \begin{matrix}sin & nx \\ √ & \overline{π} \end{matrix}] =

= \frac{1}{\sqrt{2π}} ∫_{a}^{a+2π} \frac{f(x)}{\sqrt{2π}} dx + ∑_{n=1}^{∞} [\begin{matrix}cos & nx \\ √ & \overline{π} \end{matrix} ∫_{a}^{a+2π} f(x)\begin{matrix}cos & nx \\ √ & \overline{π} \end{matrix} dx + \begin{matrix}sin & nx \\ √ & \overline{π} \end{matrix} ∫_{a}^{a+2π} f(x)\begin{matrix}sin & nx \\ √ & \overline{π} \end{matrix} dx] =

= \frac{1}{2π} ∫_{a}^{a+2π} f(x) dx + ∑_{n=1}^{∞} [\frac{cos nx}{π} ∫_{a}^{a+2π} f(x) cos nx dx + \frac{sin nx}{π} ∫_{a}^{a+2π} f(x) sin nx dx] = \frac{a_{0}}{2} + ∑_{n=1}^{∞} [a_{n} cos nx + b_{n} sin nx],

for x∈⟨a; a + 2π⟩.

beerb@frcatel.fri.utc.sk  19  http://frcatel.fri.utc.sk/˜ beerb/

\underline{Integrals}

Theorem.

The Fourier series of a function f(x)∈R_{⟨a; a+2l⟩}, a∈R, l > 0 is given by

f(x) ≈ \frac{a_{0}}{2} + ∑_{n=1}^{∞} [a_{n} cos \frac{nπx}{l} + b_{n} sin \frac{nπx}{l}], for x∈⟨a; a + 2l⟩,

where a_{0} = \frac{1}{l} ∫_{a}^{a+2l} f(x) dx, a_{n} = \frac{1}{l} ∫_{a}^{a+2l} f(x) cos \frac{nπx}{l} dx, b_{n} = \frac{1}{l} ∫_{a}^{a+2l} f(x) sin \frac{nπx}{l} dx, n∈N.

[ ⟨a; a + 2l⟩ ⟼ ⟨a; a + 2π⟩ ⟹ x ⟼ \frac{x}{2l}2π = \frac{πx}{l} ⟹ nx ⟼ \frac{nπx}{l}, n∈N]

A Fourier series converges to the function f̃(x_{0}) = \{ \begin{matrix}\frac{1}{2} [ lim_{x→x_{0}^{−}} f(x) + lim_{x→x_{0}^{+}} f(x)] , & x_{0} ∈(a; a + 2l) , \\ f(x_{0} − 2kl), & x_{0} ∈(a + 2l; a + 2l + 2kl) , \\ \frac{1}{2} [ lim_{x→a^{+}} f(x) + lim_{x→a+2l^{−}} f(x)] , & x_{0} = a + 2kl, k ∈Z. \end{matrix}

(Function f̃(x) is a periodic function with period 2l and equals to the original function at the interval ⟨a; a + 2l⟩ at points of continuity or to the average of the two limits at points of discontinuity.)

[figure: x; y; 0; a; a + 2π; ^{..................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................}^{.........................................................................................................................................................................}^{......................................................................................................................................................................................................................................................................................................................}; f; ⟨a; a + 2π⟩; x; y; 0; a − 2π; a; a + 2π; a + 4π; f̃; ^{...............................................................}; ^{..................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................}; ^{.........................................................................................................................................................................}; ^{......................................................................................................................................................................................................................................................................................................................}; ^{..................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................}; ^{.........................................................................................................................................................................}; ^{......................................................................................................................................................................................................................................................................................................................}; ⟨a; a + 2π⟩; ^{..................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................}; ^{.........................................................................................................................................................................}; ^{......................................................................................................................................................................................................................................................................................................................}; ^{..................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................} ^{..................................}; ⟨a − 2π; a⟩; ⟨a + 2π; a + 4π⟩]

[figure: x; y; 0; −l; l; ^{..................................................................................................................................................................................................................................................................................................................................................................................................................................................................}; ^{...................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................}; f; ⟨−l; l⟩; x; y; 0; −3l; −l; l; 3l; f̃; ^{...................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................}; . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .; ⟨−5l; −3l⟩; ^{..................................................................................................................................................................................................................................................................................................................................................................................................................................................................}; ^{...................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................}; ⟨−3l; −l⟩; ^{..................................................................................................................................................................................................................................................................................................................................................................................................................................................................}; ^{...................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................}; ⟨−l; l⟩; ^{..................................................................................................................................................................................................................................................................................................................................................................................................................................................................}; ^{...................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................}; ⟨l; 3l⟩; ^{..................................................................................................................................................................................................................................................................................................................................................................................................................................................................}; 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\underline{Integrals}

Theorem.

(i) If f(x) is an odd function, then a_{n} = 0, n = 0, 1, 2, … and the Fourier series collapses to Fourier Sine Series

f(x) ≈ ∑_{n=1}^{∞} b_{n} sin nx, for x∈⟨−π; π⟩, where b_{n} = \frac{2}{π} ∫_{−π}^{π} f(x) sin nx dx, n∈N.

resp. f(x) ≈ ∑_{n=1}^{∞} b_{n} sin \frac{nπx}{l}, for x∈⟨−l; l⟩, l > 0, where b_{n} = \frac{2}{l} ∫_{−l}^{l} f(x) sin \frac{nπx}{l} dx, n∈N.

(ii) If f(x) is an even function, then b_{n} = 0, n∈N and the Fourier series collapses to Fourier Cosine Series

f(x) ≈ \frac{a_{0}}{2} + ∑_{n=1}^{∞} a_{n} cos nx, for x∈⟨−π; π⟩, where a_{0} = \frac{2}{π} ∫_{−π}^{π} f(x) dx, a_{n} = \frac{2}{π} ∫_{−π}^{π} f(x) cos nx dx, n∈N.

resp. f(x) ≈ \frac{a_{0}}{2} + ∑_{n=1}^{∞} a_{n} cos \frac{nπx}{l}, for x∈⟨−l; l⟩, l > 0, where a_{0} = \frac{2}{l} ∫_{−l}^{l} f(x) dx, a_{n} = \frac{2}{l} ∫_{−l}^{l} f(x) cos \frac{nπx}{l} dx, n∈N.

x

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−l  0  l
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⟨0 ; l⟩

Function f ∈R_{⟨0 ; t⟩}

x

y

−4l 0 −3l  −2l  −l  l  2l  3l  4l

Fourier Series

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⟨−2l; −l⟩ ⟨−l; 0⟩ ⟨l; 2l⟩

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x

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−3l  −l  l  3l

Fourier Sine Series

⟨−3l; −l⟩

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⟨−l; l⟩

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⟨l; 3l⟩

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x

y

−3l 0 −l  l  3l

Fourier Cosine Series

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⟨−3l; −l⟩

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⟨l; 3l⟩

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⟨−l; l⟩

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\underline{Integrals}

Theorem.

∫_{a}^{a+2π} f(x) dx = ∫_{0}^{2π} f(x) dx = ∫_{−π}^{π} f(x) dx, for a periodic function f(x)∈R_{⟨a; a+2π⟩} with period 2π, a > 0.

∫_{a}^{a+2π} f(x) dx = ∫_{a}^{2π} f(x) dx + ∫_{2π}^{a+2π} f(x) dx = \begin{matrix}x = t − 2π, dx = dt \\ 2π ↦ 0, a + 2π ↦ a \end{matrix} = ∫_{a}^{2π} f(x) dx + ∫_{0}^{a} f(t − 2π) dt =

= f(t − 2π) = f(t) = ∫_{a}^{2π} f(x) dx + ∫_{0}^{a} f(t) dt = ∫_{a}^{2π} f(x) dx + ∫_{0}^{a} f(x) dx = ∫_{0}^{2π} f(x) dx

∫_{0}^{2π} f(x) dx = ∫_{−π}^{π} f(x) dx, for a periodic function f(x)∈R_{⟨0 ; 2π⟩} with period 2π.

∫_{a}^{b} f(x) dx = \begin{matrix}x = −t, dx = − dt \\ a ↦ −a, b ↦ −b \end{matrix} = − ∫_{−a}^{−b} f(−t) dt = ∫_{−b}^{−a} f(−t) dt = ∫_{−b}^{−a} f(−x) dx, for f(x)∈R_{⟨a; b⟩}.

∫_{a}^{b} f(x) dx = ∫_{−b}^{−a} f(−x) dx = \{ \begin{matrix}∫_{−b}^{−a} f(x) dx, & for a even function f(x)∈R_{⟨a; b⟩}, \\ − ∫_{−b}^{−a} f(x) dx, & for a odd function f(x)∈R_{⟨a; b⟩}. \end{matrix}

∫_{−a}^{a} f(x) dx = ∫_{−a}^{0} f(x) dx + ∫_{0}^{a} f(x) dx = \{ \begin{matrix}∫_{0}^{a} f(x) dx + ∫_{0}^{a} f(x) dx = 2 ∫_{0}^{a} f(x) dx, & for a even function f(x)∈R_{⟨−a; a⟩}, a > 0, \\ − ∫_{0}^{a} f(x) dx + ∫_{0}^{a} f(x) dx = 0, & for a odd function f(x)∈R_{⟨−a; a⟩}, a > 0. \end{matrix}

sin 2kπ = 0, sin kπ = 0, sin (2k + 1)\frac{π}{2} = (−1)^{k}, sin (2k − 1)\frac{π}{2} = (−1)^{k−1} = (−1)^{k+1},
cos 2kπ = 1, cos kπ = (−1)^{k}, cos (2k + 1)\frac{π}{2} = 0, for k ∈Z

∫_{0}^{2π} cos nx · cos mx dx = ∫_{−π}^{π} cos nx · cos mx dx = ∫_{a}^{a+2π} cos nx · cos mx dx = 0, for m, n∈N, m ≠ n, a∈R.

∫_{0}^{2π} cos nx · cos mx dx = u = cos nx ⇒ u′ = −n sin nx v′ = cos mx ⇒ v = \frac{sin mx}{m} = [\frac{cos nx · sin mx}{m}]_{0}^{2π} + \frac{n}{m} ∫_{0}^{2π} sin nx · sin mx dx =

= [\frac{1 · 0}{m} − \frac{1 · 0}{m}] + \frac{n}{m} ∫_{0}^{2π} sin nx · sin mx dx = \frac{n}{m} ∫_{0}^{2π} sin nx · sin mx dx = u = sin nx ⇒ u′ = n cos nx v′ = sin mx ⇒ v = −\frac{cos mx}{m} =

= \frac{n}{m} ([−\frac{sin nx · cos mx}{m}]_{0}^{2π} + \frac{n}{m} ∫_{0}^{2π} cos nx · cos mx dx) = \frac{n}{m} [−\frac{0 · 1}{m} + \frac{0 · 1}{m}]_{0}^{2π} + \frac{n^{2}}{m^{2}} ∫_{0}^{2π} cos nx · cos mx dx

⟹^{(i.e.equation)} ∫_{0}^{2π} cos nx · cos mx dx = \frac{n^{2}}{m^{2}} ∫_{0}^{2π} cos nx · cos mx dx ⟹ (1 − \frac{n^{2}}{m^{2}} ) ∫_{0}^{2π} cos nx · cos mx dx = 0

⟹^{m≠n} 1 − \frac{n^{2}}{m^{2}} = \frac{m^{2} − n^{2}}{m^{2}} ≠ 0 ⟹ ∫_{0}^{2π} cos nx · cos mx dx = 0, for m, n∈N, m ≠ n.

∫_{0}^{2π} sin nx · sin mx dx = ∫_{−π}^{π} sin nx · sin mx dx = ∫_{a}^{a+2π} sin nx · sin mx dx = 0, for m, n∈N, m ≠ n, a∈R.

∫_{0}^{2π} sin nx · sin mx dx = u = sin nx ⇒ u′ = n cos nx v′ = sin mx ⇒ v = −\frac{cos mx}{m} = [−\frac{sin nx · cos mx}{m}]_{0}^{2π} + \frac{n}{m} ∫_{0}^{2π} cos nx · cos mx dx =

= [−\frac{0 · 1}{m} + \frac{0 · 1}{m}] + \frac{n}{m} ∫_{0}^{2π} cos nx · cos mx dx = \frac{n}{m} ∫_{0}^{2π} cos nx · cos mx dx = u = cos nx ⇒ u′ = −n sin nx v′ = cos mx ⇒ v = \frac{sin mx}{m} =

= \frac{n}{m} ([\frac{cos nx · sin mx}{m}]_{0}^{2π} + \frac{n}{m} ∫_{0}^{2π} sin nx · sin mx dx) = \frac{n}{m} [\frac{1 · 0}{m} − \frac{1 · 0}{m}]_{0}^{2π} + \frac{n^{2}}{m^{2}} ∫_{0}^{2π} sin nx · sin mx dx

⟹^{(i.e.equation)} ∫_{0}^{2π} sin nx · sin mx dx = \frac{n^{2}}{m^{2}} ∫_{0}^{2π} sin nx · sin mx dx ⟹ (1 − \frac{n^{2}}{m^{2}} ) ∫_{0}^{2π} sin nx · sin mx dx = 0

⟹^{m≠n} 1 − \frac{n^{2}}{m^{2}} = \frac{m^{2} − n^{2}}{m^{2}} ≠ 0 ⟹ ∫_{0}^{2π} sin nx · sin mx dx = 0, for m, n∈N, m ≠ n.

∫_{0}^{2π} cos nx · sin mx dx = ∫_{−π}^{π} cos nx · sin mx dx = ∫_{a}^{a+2π} cos nx · sin mx dx = 0, for m, n∈N, a∈R.

∫_{0}^{2π} cos nx · sin mx dx = u = cos nx ⇒ u′ = −n sin nx v′ = sin mx ⇒ v = −\frac{cos mx}{m} = [−\frac{cos nx · cos mx}{m}]_{0}^{2π} − \frac{n}{m} ∫_{0}^{2π} sin nx · cos mx dx =

= [−\frac{1·1}{m} + \frac{1·1}{m}] − \frac{n}{m} ∫_{0}^{2π} sin nx · sin mx dx = −\frac{n}{m} ∫_{0}^{2π} sin nx · sin mx dx = u = sin nx ⇒ u′ = n cos nx v′ = cos mx ⇒ v = \frac{sin mx}{m} =

= −\frac{n}{m} ([\frac{sin nx · sin mx}{m}]_{0}^{2π} − \frac{n}{m} ∫_{0}^{2π} cos nx · sin mx dx) = −\frac{n}{m} [\frac{0 · 0}{m} − \frac{0 · 0}{m}]_{0}^{2π} + \frac{n^{2}}{m^{2}} ∫_{0}^{2π} cos nx · sin mx dx

⟹^{(i.e.equation)} ∫_{0}^{2π} cos nx · sin mx dx = \frac{n^{2}}{m^{2}} ∫_{0}^{2π} cos nx · sin mx dx ⟹ (1 − \frac{n^{2}}{m^{2}} ) ∫_{0}^{2π} cos nx · sin mx dx = 0

⟹^{m≠n} 1 − \frac{n^{2}}{m^{2}} = \frac{m^{2} − n^{2}}{m^{2}} ≠ 0 ⟹ ∫_{0}^{2π} cos nx · sin mx dx = 0, for m, n∈N, m ≠ n.

∫_{0}^{2π} cos nx · sin nx dx = \frac{1}{2} ∫_{0}^{2π} sin 2nx dx = \frac{1}{2} [−\frac{cos 2nx}{2n}]_{0}^{2π} = \frac{1}{2} [−\frac{cos 4nπ}{2n} + \frac{cos 0}{2n}] = \frac{1}{2} [−\frac{1}{2n} + \frac{1}{2n}] = 0, for m = n.

∫_{0}^{2π} cos nx dx = ∫_{−π}^{π} cos nx dx = ∫_{a}^{a+2π} cos nx dx = [\frac{sin nx}{n}]_{a}^{a+2π} = \frac{sin (na+2nπ) − sin na}{n} = \frac{sin na − sin na}{n} = 0, for n∈N, a∈R.

∫_{0}^{2π} sin nx dx = ∫_{−π}^{π} sin nx dx = ∫_{a}^{a+2π} sin nx dx = [−\frac{cos nx}{n}]_{a}^{a+2π} = \frac{cos na − cos (na+2nπ)}{n} = \frac{cos na − cos na}{n} = 0, for n∈N, a∈R.

∥1∥ = \sqrt{∫_{0}^{2π} 1^{2} dx} = \sqrt{∫_{−π}^{π} 1^{2} dx} = \sqrt{∫_{a}^{a+2π} 1^{2} dx} = \sqrt{∫_{a}^{a+2π} dx} = \sqrt{a + 2π − a} = \sqrt{2π}, for a∈R.

∥cos nx∥ = \sqrt{∫_{0}^{2π} cos^{2} nx dx} = \sqrt{∫_{−π}^{π} cos^{2} nx dx} = \sqrt{∫_{a}^{a+2π} cos^{2} nx dx} = \sqrt{π}, for n∈N, a∈R.

∫_{a}^{a+2π} cos^{2} nx dx = ∫_{a}^{a+2π} \frac{1 + cos 2nx}{2} dx = \frac{1}{2} ∫_{a}^{a+2π} dx + \frac{1}{2} ∫_{a}^{a+2π} cos 2nx dx = \frac{a + 2π − a}{2} + \frac{1}{2} · 0 = π.

∥sin nx∥ = \sqrt{∫_{0}^{2π} sin^{2} nx dx} = \sqrt{∫_{−π}^{π} sin^{2} nx dx} = \sqrt{∫_{a}^{a+2π} sin^{2} nx dx} = \sqrt{π}, for n∈N, a∈R.

∫_{a}^{a+2π} sin^{2} nx dx = ∫_{a}^{a+2π} \frac{1 − cos 2nx}{2} dx = \frac{1}{2} ∫_{a}^{a+2π} dx − \frac{1}{2} ∫_{a}^{a+2π} cos 2nx dx = \frac{a + 2π − a}{2} − \frac{1}{2} · 0 = π.

\frac{πx}{l} = t, x = \frac{lt}{π}, dx = \frac{l}{π} dt, x = a ⟹ t = \frac{πa}{l} = b, x = a + 2l ⟹ t = \frac{π(a + 2l)}{l} = \frac{πa}{l} + 2π = b + 2π

∫_{a}^{a+2l} cos \frac{nπx}{l} dx = \frac{l}{π} ∫_{b}^{b+2π} cos nt dt = \frac{l}{π} · 0 = 0, for n∈N, a∈R, l > 0.

∫_{a}^{a+2l} sin \frac{nπx}{l} dx = \frac{l}{π} ∫_{b}^{b+2π} sin nt dt = \frac{l}{π} · 0 = 0, for n∈N, a∈R, l > 0.

∫_{0}^{2π} cos \frac{nπx}{l} · cos \frac{mπx}{l} dx = \frac{l}{π} ∫_{a}^{a+2π} cos nt · cos mt dt = \frac{l · 0}{π} = 0, for m, n∈N, m ≠ n, a∈R, l > 0.

∫_{0}^{2π} sin \frac{nπx}{l} · sin \frac{mπx}{l} dx = \frac{l}{π} ∫_{a}^{a+2π} sin nt · sin mt dt = \frac{l · 0}{π} = 0, for m, n∈N, m ≠ n, a∈R, l > 0.

∫_{0}^{2π} cos \frac{nπx}{l} · sin \frac{mπx}{l} dx = \frac{l}{π} ∫_{a}^{a+2π} cos nt · sin mt dt = \frac{l · 0}{π} = 0, for m, n∈N, a∈R, l > 0.

∥1∥ = \sqrt{∫_{a}^{a+2l} 1^{2} dx} = \sqrt{[x]_{a}^{a+2l}} = \sqrt{2l}, for a∈R, l > 0.

∥cos \frac{nπx}{l}∥ = \sqrt{∫_{a}^{a+2l} cos^{2} \frac{nπx}{l} dx} = \sqrt{\frac{l}{π} ∫_{b}^{b+2π} cos^{2} nt dt} = \sqrt{\frac{l}{π} π} = \sqrt{l}, for n∈N, a∈R, l > 0.

∥sin \frac{nπx}{l}∥ = \sqrt{∫_{a}^{a+2l} sin^{2} \frac{nπx}{l} dx} = \sqrt{\frac{l}{π} ∫_{b}^{b+2π} sin^{2} nt dt} = \sqrt{\frac{l}{π} π} = \sqrt{l}, for n∈N, a∈R, l > 0.

f(x) = 1, x∈⟨0 ; 2π⟩

⟹^{Fourier series} f(x) ≈ f̃(x) = \frac{a_{0}}{2} + ∑_{n=1}^{∞} [a_{n} cos nx + b_{n} sin nx] = \frac{2}{2} + ∑_{n=1}^{∞} [0 · cos nx + 0 · sin nx] = 1 + ∑_{n=1}^{∞} 0 = 1, x∈⟨0 ; 2π⟩.

a_{0} = \frac{1}{π} ∫_{0}^{2π} dx = \frac{2π}{π} = 2, a_{n} = \frac{1}{π} ∫_{0}^{2π} cos nx dx = \frac{1}{π} [\frac{sin nx}{n}]_{0}^{2π} = \frac{1}{π} [\frac{sin 2nπ}{n} + \frac{sin 0}{n}] = \frac{1}{π} [0 − 0] = 0, n∈N.

b_{n} = \frac{1}{π} ∫_{0}^{2π} sin nx dx = \frac{1}{π} [−\frac{cos nx}{n}]_{0}^{2π} = \frac{1}{π} [−\frac{cos 2nπ}{n} + \frac{cos 0}{n}] = \frac{1}{π} [−\frac{1}{n} + \frac{1}{n}] = 0, n∈N.

[figure: x; y; 1; 0; π; 2π; ⟨0 ; 2π⟩; . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .; 1; x; y; 0; 1; −2π; 2π; 4π; f̃; . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .; ⟨−2π; 0⟩; . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .; ⟨2π; 4π⟩; . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .; . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .; ⟨0 ; 2π⟩; . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .]

f(x) = 1, x∈⟨−π; π⟩ [f(x) is a even function, i.e. b_{n} = 0, n = 1, 2, 3, …]

⟹^{Fourier Cosine series} f(x) ≈ f̃(x) = \frac{a_{0}}{2} + ∑_{n=1}^{∞} a_{n} cos nx = \frac{2}{2} + ∑_{n=1}^{∞} 0 · cos nx = 1 + ∑_{n=1}^{∞} 0 = 1, x∈⟨−π; π⟩.

a_{0} = \frac{2}{π} ∫_{0}^{π} dx = \frac{2}{π} π = 2, a_{n} = \frac{2}{π} ∫_{0}^{π} cos nx dx = \frac{2}{π} [\frac{sin nx}{n}]_{0}^{π} = \frac{2}{π} [\frac{sin nπ}{n} − \frac{sin 0}{n}] = \frac{2}{π} [\frac{0}{n} − \frac{0}{n}] = 0, n∈N.

[figure: x; y; 1; 0; −π; π; ⟨−π; π⟩; . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .; 1; x; y; 0; 1; −3π; −π; π; 3π; f̃; . . . . . . . . . . . . . . . . . . . . . . . . . . .; ⟨−3π; −π⟩; . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .; ⟨π; 3π⟩; . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .; . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .; ⟨−π; π⟩; . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .]

f(x) = 1, x∈⟨0 ; π⟩ ⟹ f_{o}(x) = \{ \begin{matrix}1, & x∈(0 ; π⟩ , \\ 0, & x = 0, \\ −1, & x∈⟨−π; 0) \end{matrix} is a odd function, i.e. a_{n} = 0, n = 0, 1, 2, …

⟹^{Fourier Sine series} f_{o}(x) ≈ f̃_{o}(x) = ∑_{n=1}^{∞} b_{n} sin nx = ∑_{n=1}^{∞} \frac{2[(−1)^{n+1} + 1] sin nx}{nπ} = ∑_{k=1}^{∞} \frac{4 sin (2k−1)x}{(2k − 1)π}, x∈⟨−π; π⟩.

⟹^{Fourier Sine series} f(x) ≈ f̃_{o}(x) = ∑_{k=1}^{∞} \frac{4 sin (2k−1)x}{(2k − 1)π} = \frac{4}{π} ∑_{k=1}^{∞} \frac{sin (2k−1)x}{2k − 1}, x∈⟨0 ; π⟩.

b_{n} = \frac{2}{π} ∫_{0}^{π} f_{o}(x) sin nx dx = \frac{2}{π} ∫_{0}^{π} sin nx dx = \frac{2}{π} [−\frac{cos nx}{n}]_{0}^{π} = \frac{2}{π} [−\frac{cos nπ}{n} + \frac{cos 0}{n}] = \frac{2}{π} [−\frac{(−1)^{n}}{n} + \frac{1}{n}] =

= \frac{2}{π} \frac{(−1)^{n+1} + 1}{n} = \frac{2[(−1)^{n+1} + 1]}{nπ} = \{ \begin{matrix}\frac{2(1 + 1)}{nπ} = \frac{4}{nπ} = \frac{4}{(2k − 1)π} , & for n = 2k − 1, \\ \frac{2(−1 + 1)}{nπ} = 0, & for n = 2k, k ∈N. \end{matrix}

[figure: x; y; 0; −1; 1; −π; π; ⟨−π; π⟩; . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .; 1; x; y; 0; 1; −1; −3π; −π; π; 3π; f̃; . . . . . . . . . . . . . . . . . . . . . . . . . . .; ⟨−3π; −π⟩; . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .; . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .; ⟨π; 3π⟩; . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .; . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .; . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .; f_{o}; ⟨−π; π⟩; . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .; . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .]

f(x) = x, x∈⟨0 ; 2π⟩

⟹^{Fourier series} f(x) ≈ f̃(x) = \frac{a_{0}}{2} + ∑_{n=1}^{∞} [a_{n} cos nx + b_{n} sin nx] = π + ∑_{n=1}^{∞} [0 · cos nx − \frac{2}{n} sin nx] = π − 2 ∑_{n=1}^{∞} \frac{sin nx}{n}, x∈⟨0 ; 2π⟩.

a_{0} = \frac{1}{π} ∫_{0}^{2π} x dx = \frac{1}{π} [\frac{x^{2}}{2} ]_{0}^{2π} dx = \frac{4π^{2}}{2π} = 2π,

a_{n} = \frac{1}{π} ∫_{0}^{2π} x cos nx dx =_{73}^{page} \frac{1}{π} [\frac{x sin nx}{n} + \frac{cos nx}{n^{2}}]_{0}^{2π} = \frac{1}{π} [\frac{2π · 0}{n} + \frac{1}{n^{2}} − \frac{0 · 0}{n} − \frac{1}{n^{2}} ] = 0, n∈N,

b_{n} = \frac{1}{π} ∫_{0}^{2π} x sin nx dx =_{72}^{page} \frac{1}{π} [−\frac{x cos nx}{n} + \frac{sin nx}{n^{2}}]_{0}^{2π} = \frac{1}{π} [−\frac{2π · 1}{n} + \frac{0}{n^{2}} + \frac{0 · 1}{n} − \frac{0}{n^{2}} ] = −\frac{2}{n}, n∈N.

[figure: x; y; π; 2π; 0; π; 2π; ⟨0 ; 2π⟩; ^{...............................................................................................................................................................................................................................................................................................................................................}; x; x; y; 0; π; 2π; −2π; 2π; 4π; f̃; ^{..............................................................................................................................................................................................................}; ⟨−2π; 0⟩; ^{...............................................................................................................................................................................................................................................................................................................................................}; ⟨2π; 4π⟩; ^{...............................................................................................................................................................................................................................................................................................................................................}; ^{...........................................}; ⟨0 ; 2π⟩; ^{...............................................................................................................................................................................................................................................................................................................................................}]

f(x) = x, x∈⟨0 ; π⟩ ⟹ f_{e}(x) = |x| = \{ \begin{matrix}x, & x∈(0 ; π⟩ , \\ −x, & x∈⟨−π; 0) \end{matrix} is a even function, b_{n} = 0, n = 1, 2, 3, …

⟹^{Fourier Cosine series} f_{e}(x) ≈ f̃_{e}(x) = \frac{a_{0}}{2} + ∑_{n=1}^{∞} a_{n} cos nx = \frac{π}{2} + ∑_{n=1}^{∞} \frac{2[(−1)^{n} − 1] cos nx}{n^{2}π} = \frac{π}{2} − ∑_{k=1}^{∞} \frac{4 cos (2k−1)π}{(2k − 1)^{2}π}, x∈⟨−π; π⟩.

⟹^{Fourier Cosine series} f(x) ≈ f̃_{e}(x) = \frac{π}{2} − ∑_{k=1}^{∞} \frac{4 cos (2k−1)π}{(2k − 1)^{2}π} = \frac{π}{2} − 4 ∑_{k=1}^{∞} \frac{cos (2k−1)π}{(2k − 1)^{2}π}, x∈⟨0 ; π⟩.

a_{0} = \frac{2}{π} ∫_{0}^{π} f_{e}(x) dx = \frac{2}{π} ∫_{0}^{π} |x| dx = \frac{2}{π} ∫_{0}^{π} x dx = \frac{2}{π} [\frac{x^{2}}{2} ]_{0}^{π} = \frac{2}{π} \frac{π^{2}}{2} = π,

a_{n} = \frac{2}{π} ∫_{0}^{π} f_{e}(x) cos nx dx = \frac{2}{π} ∫_{0}^{π} x cos nx dx = \frac{2}{π} [\frac{x sin nx}{n} + \frac{cos nx}{n^{2}}]_{0}^{π} = \frac{2}{π} [\frac{π sin nπ}{n} + \frac{cos nπ}{n^{2}} − \frac{0 · sin 0}{n} − \frac{cos 0}{n^{2}}] =

= \frac{2}{π} [0 + \frac{(−1)^{n}}{n^{2}} − 0 − \frac{1}{n^{2}} ] = \frac{2}{π} \frac{(−1)^{n} − 1}{n^{2}} = \{ \begin{matrix}\frac{2(−1 − 1)}{n^{2}π} = \frac{−4}{n^{2}π} = \frac{−4}{(2k − 1)^{2}π} , & for n = 2k − 1, \\ \frac{2(1 − 1)}{n^{2}π} = 0, & for n = 2k, k ∈N. \end{matrix}

[figure: x; y; π; 0; −π; π; ⟨0 ; π⟩; ^{........................................................................................................................................................................}; x; x; y; π; 0; −3π; −π; π; 3π; f̃; ^{.......................................}; ⟨−3π; −π⟩; . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .; ^{........................................................................................................................................................................}; ⟨π; 3π⟩; . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .; ^{........................................................................................................................................................................}; . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .; ^{...........................................}; ⟨−π; π⟩; . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .; ^{........................................................................................................................................................................}; f_{e}]

f(x) = x, x∈⟨−π; π⟩ [f(x) is a odd function, i.e. a_{n} = 0, n = 0, 1, 2, …]

⟹^{Fourier Sine series} f(x) ≈ f̃(x) = ∑_{n=1}^{∞} b_{n} sin nx = ∑_{n=1}^{∞} \frac{2(−1)^{n+1}}{n} sin nx = 2 ∑_{n=1}^{∞} \frac{(−1)^{n+1} sin nx}{n}, x∈⟨−π; π⟩.

b_{n} = \frac{1}{π} ∫_{−π}^{π} x sin nx dx = \frac{1}{π} [−\frac{x cos nx}{n} + \frac{sin nx}{n^{2}}]_{−π}^{π} = \frac{1}{π} [−\frac{π cos nπ}{n} + \frac{sin nπ}{n^{2}} + \frac{−π cos [−nπ]}{n} − \frac{sin [−nπ]}{n^{2}}] =

= \frac{1}{π} [−\frac{π cos nπ}{n} + \frac{0}{n^{2}} − \frac{π cos nπ}{n} − \frac{0}{n^{2}} ] = −\frac{2 cos nπ}{n} = −\frac{2(−1)^{n}}{n} = \frac{2(−1)^{n+1}}{n}, n∈N,

resp. b_{n} = \frac{2}{π} ∫_{0}^{π} x sin nx dx = \frac{2}{π} [−\frac{x cos nx}{n} + \frac{sin nx}{n^{2}}]_{0}^{π} = \frac{2}{π} [−\frac{π(−1)^{n}}{n} + \frac{0}{n^{2}} + \frac{0 · 1}{n} − \frac{0}{n^{2}} ] = \frac{2(−1)^{n+1}}{n}, n∈N.

[figure: x; y; −π; π; 0; −π; π; ⟨−π; π⟩; ^{...............................................................................................................................................................................................................................................................................................................................................}; x; x; y; −π; π; 0; −3π; −π; π; 3π; f̃; ^{.......................................}; ⟨−3π; −π⟩; ^{...............................................................................................................................................................................................................................................................................................................................................}; ⟨π; 3π⟩; ^{...............................................................................................................................................................................................................................................................................................................................................}; ^{..................................................................................................................................................................................................................}; ⟨−π; π⟩; ^{...............................................................................................................................................................................................................................................................................................................................................}]

f̃(x) = 2 ∑_{n=1}^{∞} \frac{(−1)^{n+1} sin nx}{n}, x∈⟨−π; π⟩,  f̃(x) = f(x) = x, x∈(−π; π), f̃(−π) = f̃(π) = \frac{f(−π) + f(π)}{2} = \frac{−π + π}{2} = 0.

f (\frac{π}{2} ) = \frac{π}{2} ,

f̃(\frac{π}{2} ) = 2 ∑_{n=1}^{∞} \frac{(−1)^{n+1}}{n} sin \frac{nπ}{2} = 2 [sin \frac{π}{2} − \frac{1}{2} sin π + \frac{1}{3} sin \frac{3π}{2} − \frac{1}{4} sin 2π + \frac{1}{5} sin \frac{5π}{2} − \frac{1}{6} sin 3π + ⋯] =

= 2 [1 − \frac{1}{2} · 0 + \frac{1}{3}(−1) − \frac{1}{4} · 0 + \frac{1}{5} · 1 − \frac{1}{6} · 0 + ⋯] = 2 [1 − \frac{1}{3} + \frac{1}{5} − \frac{1}{7} + \frac{1}{9} + ⋯] = 2 ∑_{k=1}^{∞} \frac{(−1)^{k+1}}{2k − 1} ,

resp. f̃(\frac{π}{2} ) = 2 ∑_{n=1}^{∞} \frac{(−1)^{n+1}}{n} sin \frac{nπ}{2} = 2 ∑_{k=1}^{∞} [\frac{(−1)^{(2k−1)+1}}{2k − 1} sin \frac{(2k−1)π}{2} + \frac{(−1)^{2k+1}}{2k} sin \frac{2kπ}{2} ] =

= 2 ∑_{k=1}^{∞} [\frac{(−1)^{2k}}{2k − 1} sin \frac{(2k−1)π}{2} + \frac{(−1)^{2k+1}}{2k} sin kπ] = 2 ∑_{k=1}^{∞} [\frac{1}{2k − 1} · (−1)^{k+1} − \frac{1}{2k} · 0] = 2 ∑_{k=1}^{∞} \frac{(−1)^{k+1}}{2k − 1} .

\frac{π}{2} = f (\frac{π}{2} ) = f̃(\frac{π}{2} ) = 2 ∑_{k=1}^{∞} \frac{(−1)^{k+1}}{2k − 1} ⟹ \frac{π}{4} = ∑_{k=1}^{∞} \frac{(−1)^{k+1}}{2k − 1} ⟹ \frac{π}{4} = ∑_{n=1}^{∞} \frac{(−1)^{n+1}}{2n − 1} = 1 − \frac{1}{3} + \frac{1}{5} − \frac{1}{7} + ⋯ + \frac{(−1)^{n+1}}{2n − 1} + · · ·.

f(x) = x^{2}, x∈⟨0 ; 2π⟩

⟹^{Fourier series} f(x) ≈ f̃(x) = \frac{a_{0}}{2} + ∑_{n=1}^{∞} [a_{n} cos nx + b_{n} sin nx] = \frac{4π^{2}}{3} + ∑_{n=1}^{∞} [\frac{4 cos nx}{n^{2}} − \frac{4π sin nx}{n}], x∈⟨0 ; 2π⟩.

a_{0} = \frac{1}{π} ∫_{0}^{2π} x^{2} dx = \frac{1}{π} [\frac{x^{3}}{3} ]_{0}^{2π} = \frac{1}{π} [\frac{(2π)^{3}}{3} − \frac{0}{3}] = \frac{8π^{3}}{3π} = \frac{8π^{2}}{3} ,

a_{n} = \frac{1}{π} ∫_{0}^{2π} x^{2} cos nx dx =_{73}^{page} \frac{1}{π} [\frac{x^{2} sin nx}{n} + \frac{2x cos nx}{n^{2}} − \frac{2 sin nx}{n^{3}}]_{0}^{2π} = \frac{1}{π} [\frac{4π^{2} sin 2nπ}{n} + \frac{4π cos 2nπ}{n^{2}} − \frac{2 sin 2nπ}{n^{3}} −

−\frac{0 · sin 0}{n} − \frac{0 · cos 0}{n^{2}} + \frac{2 sin 0}{n^{3}}] = \frac{1}{π} [0 + \frac{4π}{n^{2}} − 0 − 0 − 0 + 0] = \frac{4}{n^{2}} , n∈N,

b_{n} = \frac{1}{π} ∫_{0}^{2π} x^{2} sin nx dx =_{72}^{page} \frac{1}{π} [−\frac{x^{2} cos nx}{n} + \frac{2x sin nx}{n^{2}} + \frac{2 cos nx}{n^{3}}]_{0}^{2π} = \frac{1}{π} [−\frac{4π^{2} cos 2nπ}{n} + \frac{4π sin 2nπ}{n^{2}} + \frac{2 cos 2nπ}{n^{3}} +

+\frac{0 · cos 0}{n} − \frac{0 · sin 0}{n^{2}} − \frac{2 cos 0}{n^{3}}] = \frac{1}{π} [−\frac{4π^{2}}{n} + 0 + \frac{2}{n^{3}} + 0 − 0 − \frac{2}{n^{3}} ] = −\frac{4π}{n}, n∈N.

[figure: x; y; 0; −2π; 2π; 2π^{2}; 4π^{2}; x^{2}; ^{...................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................}; ⟨0 ; 2π⟩; x; y; 0; −4π; −2π; 2π; 4π; 2π^{2}; 4π^{2}; ^{...............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................}; ^{...................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................}; ⟨−4π; −2π⟩; ^{...................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................}; ⟨−2π; 0⟩; ^{...................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................}; ⟨2π; 4π⟩; ^{............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................}; f̃; ^{...................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................}; ⟨0 ; 2π⟩]

f(x) = x^{2}, x∈⟨0 ; π⟩ ⟹ f_{o}(x) = x · |x| = \{ \begin{matrix}x^{2}, & x∈(0 ; π⟩ , \\ −x^{2}, & x∈⟨−π; 0) \end{matrix} is a odd function, a_{n} = 0, n = 0, 1, 2, …

⟹^{Fourier Sine series} f_{o}(x) ≈ f̃(x) = ∑_{n=1}^{∞} b_{n} sin nx = ∑_{n=1}^{∞} [\frac{2π(−1)^{n+1}}{n} + \frac{4[(−1)^{n} − 1]}{n^{3}π}] sin nx, x∈⟨−π; π⟩.

⟹^{Fourier Sine series} f(x) ≈ f̃(x) = ∑_{n=1}^{∞} [\frac{2π(−1)^{n+1}}{n} + \frac{4[(−1)^{n} − 1]}{n^{3}π}] sin nx = \frac{2}{π} ∑_{n=1}^{∞} \frac{2(−1)^{n} − (−1)^{n}π^{2}n^{2} − 2}{n^{3}} sin nx, x∈⟨0 ; π⟩.

b_{n} = \frac{2}{π} ∫_{0}^{π} f_{o}(x) sin nx dx = \frac{2}{π} ∫_{0}^{π} x^{2} sin nx dx = \frac{2}{π} [−\frac{x^{2} cos nx}{n} + \frac{2x sin nx}{n^{2}} + \frac{2 cos nx}{n^{3}}]_{0}^{π} =

= \frac{2}{π} [−\frac{π^{2} cos nπ}{n} + \frac{2π sin nπ}{n^{2}} + \frac{2 cos nπ}{n^{3}} + \frac{0 · cos 0}{n} − \frac{0 · sin 0}{n^{2}} − \frac{2 cos 0}{n^{3}}] =

= \frac{2}{π} [−\frac{π^{2}(−1)^{n}}{n} + 0 + \frac{2(−1)^{n}}{n^{3}} + 0 − 0 − \frac{2}{n^{3}} ] = \frac{2π(−1)^{n+1}}{n} + \frac{4[(−1)^{n} − 1]}{n^{3}π} = \frac{2}{π} \frac{2(−1)^{n} − (−1)^{n}π^{2}n^{2} − 2}{n^{3}}, n∈N.

[figure: x; y; 0; −π; π; π^{2}; −π^{2}; x^{2}; ^{....................................................................................................................................................................................................................................................................}; ^{...................................................................................................................................................................................................................................................................}; ⟨−π; π⟩; x; y; 0; −5π; −3π; −π; π; 3π; 5π; π^{2}; −π^{2}; ^{......................................................................................................................................................................................................................}; ^{....................................................................................................................................................................................................................................................................}; ^{...................................................................................................................................................................................................................................................................}; ⟨−5π; −3π⟩; ^{....................................................................................................................................................................................................................................................................}; ^{...................................................................................................................................................................................................................................................................}; ⟨−3π; −π⟩; ^{....................................................................................................................................................................................................................................................................}; ^{...................................................................................................................................................................................................................................................................}; ⟨π; 3π⟩; ^{....................................................................................................................................................................................................................................................................}; ^{...................................................................................................................................................................................................................................................................}; ⟨3π; 5π⟩; f̃; ^{......................................................................................................................................................................................................................}; ^{....................................................................................................................................................................................................................................................................}; ^{...................................................................................................................................................................................................................................................................}; ⟨−π; π⟩]

f(x) = x^{2}, x∈⟨−π; π⟩ [f(x) is a even function, b_{n} = 0, n = 1, 2, 3, …]

⟹^{Fourier Cosine series} f(x) ≈ f̃(x) = \frac{a_{0}}{2} + ∑_{n=1}^{∞} a_{n} cos nx = \frac{π^{2}}{3} + ∑_{n=1}^{∞} \frac{4(−1)^{n} cos nx}{n^{2}} = \frac{π^{2}}{3} + 4 ∑_{n=1}^{∞} \frac{(−1)^{n} cos nx}{n^{2}}, x∈⟨−π; π⟩.

a_{0} = \frac{2}{π} ∫_{0}^{π} x^{2} dx = \frac{2}{π} [\frac{x^{3}}{3} ]_{0}^{π} = \frac{2}{π} [\frac{π^{3}}{3} − \frac{0}{3}] = \frac{2π^{3}}{3π} = \frac{2π^{2}}{3} ,

a_{n} = \frac{2}{π} ∫_{0}^{π} x^{2} cos nx dx = \frac{2}{π} [\frac{x^{2} sin nx}{n} + \frac{2x cos nx}{n^{2}} − \frac{2 sin nx}{n^{3}}]_{0}^{π} = \frac{2}{π} [\frac{π^{2} sin nπ}{n} + \frac{2π cos nπ}{n^{2}} − \frac{2 sin nπ}{n^{3}} −

−\frac{0 · sin 0}{n} − \frac{0 · cos 0}{n^{2}} + \frac{2 sin 0}{n^{3}}] = \frac{2}{π} [0 + \frac{2π(−1)^{n}}{n^{2}} − 0 − 0 − 0 + 0] = \frac{4(−1)^{n}}{n^{2}} , n∈N.

[figure: x; y; 0; −π; π; π^{2}; x^{2}; ^{....................................................................................................................................................................................................................................................................}; ^{...................................................................................................................................................................................................................................................................}; ⟨−π; π⟩; x; y; 0; −5π; −3π; −π; π; 3π; 5π; π^{2}; ^{......................................................................................................................................................................................................................}; ^{....................................................................................................................................................................................................................................................................}; ^{...................................................................................................................................................................................................................................................................}; ⟨−5π; −3π⟩; ^{....................................................................................................................................................................................................................................................................}; ^{...................................................................................................................................................................................................................................................................}; ⟨−3π; −π⟩; ^{....................................................................................................................................................................................................................................................................}; ^{...................................................................................................................................................................................................................................................................}; ⟨π; 3π⟩; ^{....................................................................................................................................................................................................................................................................}; ^{...................................................................................................................................................................................................................................................................}; ⟨3π; 5π⟩; f̃; . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .; ^{....................................................................................................................................................................................................................................................................}; ^{...................................................................................................................................................................................................................................................................}; f_{e}; ⟨−π; π⟩]

f̃(x) = \frac{π^{2}}{3} + 4 ∑_{n=1}^{∞} \frac{(−1)^{n} cos nx}{n^{2}}, x∈⟨−π; π⟩,  f̃(x) = f(x) = x^{2}, x∈⟨−π; π⟩.

f(0) = 0, f̃(0) = \frac{π^{2}}{3} + 4 ∑_{n=1}^{∞} \frac{(−1)^{n} cos (0 · n)}{n^{2}} = \frac{π^{2}}{3} + 4 ∑_{n=1}^{∞} \frac{(−1)^{n} · 1}{n^{2}} = \frac{π^{2}}{3} + 4 ∑_{n=1}^{∞} \frac{(−1)^{n}}{n^{2}} ,

0 = f(0) = f̃(0) = \frac{π^{2}}{3} + 4 ∑_{n=1}^{∞} \frac{(−1)^{n}}{n^{2}} ⟹ \frac{π^{2}}{3} = −4 ∑_{n=1}^{∞} \frac{(−1)^{n}}{n^{2}} ⟹ \frac{π^{2}}{12} = ∑_{n=1}^{∞} \frac{(−1)^{n+1}}{n^{2}} = 1 − \frac{1}{4} + \frac{1}{9} − \frac{1}{16} + ⋯ + \frac{(−1)^{n+1}}{n^{2}} + · · ·.

f(π) = π^{2}, f̃(π) = \frac{π^{2}}{3} + 4 ∑_{n=1}^{∞} \frac{(−1)^{n} cos nx}{n^{2}} = \frac{π^{2}}{3} + 4 ∑_{n=1}^{∞} \frac{(−1)^{n}(−1)^{n}}{n^{2}} = \frac{π^{2}}{3} + 4 ∑_{n=1}^{∞} \frac{1}{n^{2}} ,

π^{2} = f(π) = f̃(π) = \frac{π^{2}}{3} + 4 ∑_{n=1}^{∞} \frac{1}{n^{2}} ⟹ \frac{2π^{2}}{3} = 4 ∑_{n=1}^{∞} \frac{1}{n^{2}} ⟹ \frac{π^{2}}{6} = ∑_{n=1}^{∞} \frac{1}{n^{2}} = 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + ⋯ + \frac{1}{n^{2}} + · · ·.

f(x) = x, F (x) =_{page28}^{Fourier (Sine) series} 2 ∑_{n=1}^{∞} \frac{(−1)^{n+1} sin nx}{n}, x∈⟨−π; π⟩.

g(x) = x^{2}, G(x) =_{page30}^{Fourier (Cosine) series} \frac{π^{2}}{3} + 4 ∑_{n=1}^{∞} \frac{(−1)^{n} cos nx}{n^{2}}, x∈⟨−π; π⟩.

∀x∈R: [x^{2}]′ = 2x, x^{2} = ∫_{0}^{x} 2t dt

⟹ ∀x∈(−π; π) : f(x) = \frac{g′(x)}{2}, F (x) = \frac{G′(x)}{2}, g(x) = 2 ∫_{0}^{x} f(t) dt, G(x) = 2 ∫_{0}^{x} F (t) dt.

G(x) = 2 ∫_{0}^{x} F (t) dt = 2 ∫_{0}^{x} [2 ∑_{n=1}^{∞} \frac{(−1)^{n+1} sin nt}{n}] dt = 4 ∑_{n=1}^{∞} ∫_{0}^{x} \frac{(−1)^{n+1} sin nt}{n} dt = 4 ∑_{n=1}^{∞} \frac{(−1)^{n+1}}{n} ∫_{0}^{x} sin nt dt =

= 4 ∑_{n=1}^{∞} \frac{(−1)^{n+1}}{n} [\frac{− cos nt}{n}]_{0}^{x} = 4 ∑_{n=1}^{∞} \frac{(−1)^{n+1}(− cos nx + 1)}{n^{2}} = 4 ∑_{n=1}^{∞} \frac{(−1)^{n+1}}{n^{2}} − 4 ∑_{n=1}^{∞} \frac{(−1)^{n+1} cos nx}{n^{2}}

⟹^{Fourier series} 4 ∑_{n=1}^{∞} \frac{(−1)^{n+1}}{n^{2}} = \frac{a_{0}}{2} ⟹ \frac{a_{0}}{2} = \frac{1}{2} \frac{2}{π} ∫_{0}^{π} g(x) dx = \frac{1}{π} ∫_{0}^{π} x^{2} dx = \frac{1}{π} [\frac{x^{3}}{3} ]_{0}^{π} dx = \frac{1}{π} \frac{π^{3}}{3} = \frac{π^{2}}{3}

⟹ 4 ∑_{n=1}^{∞} \frac{(−1)^{n+1}}{n^{2}} = \frac{π^{2}}{3} ⟹ ∑_{n=1}^{∞} \frac{(−1)^{n+1}}{n^{2}} = \frac{π^{2}}{12}.

⟹^{Fourier (Cosine) series} g(x) = x^{2}, G(x) = \frac{π^{2}}{3} + 4 ∑_{n=1}^{∞} \frac{(−1)^{n} cos nx}{n^{2}}, x∈⟨−π; π⟩.

F (x) = \frac{G′(x)}{2} = \frac{1}{2} [\frac{π^{2}}{3} + 4 ∑_{n=1}^{∞} \frac{(−1)^{n} cos nx}{n^{2}}]′ = \frac{1}{2} · 4 [∑_{n=1}^{∞} \frac{(−1)^{n} cos nx}{n^{2}}]′ = 2 ∑_{n=1}^{∞} [\frac{(−1)^{n} cos nx}{n^{2}}]′ =

= 2 ∑_{n=1}^{∞} \frac{(−1)^{n}}{n^{2}} [cos nx]′ = 2 ∑_{n=1}^{∞} \frac{(−1)^{n}(−n) sin nx}{n^{2}} = 2 ∑_{n=1}^{∞} \frac{(−1)^{n+1} sin nx}{n}

⟹^{Fourier (Sine) series} f(x) = x, F (x) = 2 ∑_{n=1}^{∞} \frac{(−1)^{n+1} sin nx}{n}, x∈⟨−π; π⟩.

\underline{Integrals}

Integration and Series

e^{x} = ∑_{k=0}^{∞} \frac{x^{k}}{k!} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + ⋯ + \frac{x^{k}}{k!} + ⋯ ⇉ for x∈R,

e^{−x} = ∑_{k=0}^{∞} \frac{(−x)^{k}}{k!} = ∑_{k=0}^{∞} \frac{(−1)^{k}x^{k}}{k!} ⇉ , e^{(−x^{2})} = ∑_{k=0}^{∞} \frac{(−x^{2})^{k}}{k!} = ∑_{k=0}^{∞} \frac{(−1)^{k}x^{2k}}{k!} ⇉ , e^{(x^{2})} = ∑_{k=0}^{∞} \frac{(x^{2})^{k}}{k!} = ∑_{k=0}^{∞} \frac{x^{2k}}{k!} ⇉ for x∈R.

∫ e^{x} dx = ∫ [∑_{k=0}^{∞} \frac{x^{k}}{k!} ] dx = ∑_{k=0}^{∞} ∫ \frac{x^{k}}{k!} dx = ∑_{k=0}^{∞} \frac{x^{k+1}}{(k + 1)k!} + c_{1} = ∑_{k=0}^{∞} \frac{x^{k+1}}{(k + 1)!} + c_{1} = \begin{matrix}k + 1 = n \\ k = 0, n = 1 \end{matrix} = c_{1} + ∑_{n=1}^{∞} \frac{x^{n}}{n!} = 1 = \frac{x^{0}}{0!} =

= c_{1} − 1 + \frac{x^{0}}{0!} + ∑_{n=1}^{∞} \frac{x^{n}}{n!} = c_{1} − 1 = c = c + ∑_{n=0}^{∞} \frac{x^{n}}{n!} = e^{x} +c, for x∈R.

∫ e^{x} dx = ∫ [1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + ⋯ + \frac{x^{k}}{k!} + ⋯] dx = ∫ dx + ∫ x dx + ∫ \frac{x^{2}}{2!} dx + ∫ \frac{x^{3}}{3!} dx + ⋯ + ∫ \frac{x^{k}}{k!} dx + ⋯ =

= c_{1} + x + \frac{x^{2}}{2} + \frac{x^{3}}{3 · 2!} + \frac{x^{4}}{4 · 3!} + ⋯ + \frac{x^{k+1}}{(k + 1)k!} + ⋯ = c_{1} − 1 + 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \frac{x^{4}}{4!} + ⋯ + \frac{x^{k+1}}{(k + 1)!} + ⋯ = e^{x} +c, for x∈R.

∫ e^{−x} dx = ∫ [∑_{k=0}^{∞} \frac{(−1)^{k}x^{k}}{k!} ] dx = ∑_{k=0}^{∞} ∫ \frac{(−1)^{k}x^{k}}{k!} dx = ∑_{k=0}^{∞} \frac{(−1)^{k}x^{k+1}}{(k + 1)k!} + c_{1} = \begin{matrix}k + 1 = n \\ k = 0, n = 1 \end{matrix} = c_{1} + ∑_{n=1}^{∞} \frac{(−1)^{n−1}x^{n}}{n!} =

= c_{1} − ∑_{n=1}^{∞} \frac{(−1)^{n}x^{n}}{n!} = 1 = \frac{x^{0}}{0!} = c_{1} + 1 − [\frac{x^{0}}{0!} + ∑_{n=1}^{∞} \frac{(−1)^{n}x^{n}}{n!} ] = c_{1} + 1 = c = c − ∑_{n=0}^{∞} \frac{(−1)^{n}x^{n}}{n!} = − e^{−x} +c, for x∈R.

∫ e^{−x} dx = ∫ [1 − x + \frac{x^{2}}{2!} − \frac{x^{3}}{3!} + ⋯ + \frac{(−1)^{k}x^{k}}{k!} + ⋯] dx = ∫ dx − ∫ x dx + ∫ \frac{x^{2}}{2!} dx − ∫ \frac{x^{3}}{3!} dx + ⋯ + ∫ \frac{(−1)^{k}x^{k}}{k!} dx + ⋯ =

= c_{1} + x − \frac{x^{2}}{2} + \frac{x^{3}}{3·2!} − \frac{x^{4}}{4·3!} + ⋯ + \frac{(−1)^{k}x^{k+1}}{(k + 1)k!} + ⋯ = c_{1}+1−[1 − x + \frac{x^{2}}{2!} − \frac{x^{3}}{3!} + ⋯ + \frac{(−1)^{k+1}x^{k+1}}{(k + 1)!} + ⋯] = − e^{x} +c,

for x∈R.

∫ e^{(−x^{2})} dx = ∫ [∑_{k=0}^{∞} \frac{(−1)^{k}x^{2k}}{k!} ] dx = ∑_{k=0}^{∞} ∫ \frac{(−1)^{k}x^{2k}}{k!} dx = ∑_{k=0}^{∞} \frac{(−1)^{k}x^{2k+1}}{(2k + 1)k!} + c, for x∈R.

∫ e^{(−x^{2})} dx = ∫ [1 − x^{2} + \frac{x^{4}}{2!} − \frac{x^{6}}{3!} + ⋯ + \frac{(−1)^{k}x^{2k}}{k!} + ⋯] dx = ∫ dx−∫ x^{2} dx+∫ \frac{x^{4}}{2!} dx−∫ \frac{x^{6}}{3!} dx+⋯+∫ \frac{(−1)^{k}x^{2k}}{k!} dx+⋯ =

= x − \frac{x^{3}}{3} + \frac{x^{5}}{5 · 2!} − \frac{x^{7}}{7 · 3!} + ⋯ + \frac{(−1)^{k}x^{2k+1}}{(2k + 1)k!} + ⋯ + c = ∑_{k=0}^{∞} \frac{(−1)^{k}x^{2k+1}}{(2k + 1)k!} + c, for x∈R.

∫ e^{(x^{2})} dx = ∫ [∑_{k=0}^{∞} \frac{kx^{2k}}{k!} ] dx = ∑_{k=0}^{∞} ∫ \frac{x^{2k}}{k!} dx = ∑_{k=0}^{∞} \frac{x^{2k+1}}{(2k + 1)k!} + c, for x∈R.

∫ e^{(x^{2})} dx = ∫ [1 + x^{2} + \frac{x^{4}}{2!} + \frac{x^{6}}{3!} + ⋯ + \frac{x^{2k}}{k!} + ⋯] dx = ∫ dx + ∫ x^{2} dx + ∫ \frac{x^{4}}{2!} dx + ∫ \frac{x^{6}}{3!} dx + ⋯ + ∫ \frac{x^{2k}}{k!} dx + ⋯ =

= x + \frac{x^{3}}{3} + \frac{x^{5}}{5 · 2!} + \frac{x^{7}}{7 · 3!} + \frac{x^{9}}{9 · 4!} + ⋯ + \frac{x^{2k+1}}{(2k + 1)k!} + ⋯ + c = ∑_{k=0}^{∞} \frac{x^{2k+1}}{(2k + 1)k!} + c, for x∈R.

sin x = ∑_{k=0}^{∞} \frac{(−1)^{k}x^{2k+1}}{(2k + 1)!} = x − \frac{x^{3}}{3!} + \frac{x^{5}}{5!} − \frac{x^{7}}{7!} + ⋯ + \frac{(−1)^{k}x^{2k+1}}{(2k + 1)!} + ⋯ ⇉ for x∈R, \frac{sin x}{x} = ∑_{k=0}^{∞} \frac{(−1)^{k}x^{2k}}{(2k + 1)!} ⇉ for x∈R,

cos x = ∑_{k=0}^{∞} \frac{(−1)^{k}x^{2k}}{(2k)!} = 1 − \frac{x^{2}}{2!} + \frac{x^{4}}{4!} − \frac{x^{6}}{6!} + ⋯ + \frac{(−1)^{k}x^{2k}}{(2k)!} + ⋯ ⇉ for x∈R, \frac{cos x}{x} = ∑_{k=0}^{∞} \frac{(−1)^{k}x^{2k−1}}{(2k)!} ⇉ for x∈R − \{0\}.

∫ sin x dx = ∫ [∑_{k=0}^{∞} \frac{(−1)^{k}x^{2k+1}}{(2k + 1)!} ] dx =∑_{k=0}^{∞} ∫ \frac{(−1)^{k}x^{2k+1}}{(2k + 1)!} dx =∑_{k=0}^{∞} \frac{(−1)^{k}x^{2k+2}}{(2k+2)(2k+1)!} +c_{1} = \begin{matrix}k + 1 = n, k = 0 \\ 2k+2 = 2n, n = 1 \end{matrix} =∑_{n=1}^{∞} \frac{(−1)^{n−1}x^{2n}}{(2n)!} +c_{1} =

= 1 = \frac{(−1)^{0}x^{2·0}}{(2 · 0)!} = c_{1} + 1 − \frac{(−1)^{0}x^{2·0}}{(2 · 0)!} − ∑_{n=1}^{∞} \frac{(−1)^{n}x^{2n}}{(2n)!} = c_{1} + 1 = c = c − ∑_{n=0}^{∞} \frac{(−1)^{n}x^{2n}}{(2n)!} = − cos x + c, for x∈R.

∫ sin x dx = ∫ [x − \frac{x^{3}}{3!} + \frac{x^{5}}{5!} − ⋯ + \frac{(−1)^{k}x^{2k+1}}{(2k + 1)!} + …] dx = ∫ x dx − ∫ \frac{x^{3}}{3!} dx + ∫ \frac{x^{5}}{5!} dx − ⋯ + ∫ \frac{(−1)^{k}x^{2k+1}}{(2k + 1)!} dx + ⋯ =

= \frac{x^{2}}{2} − \frac{x^{4}}{4 · 3!} + \frac{x^{6}}{6 · 5!} −⋯+ \frac{(−1)^{k}x^{2k+2}}{(2k + 2)(2k + 1)!} +⋯+c_{1} = −1+ \frac{x^{2}}{2!} − \frac{x^{4}}{4!} + \frac{x^{6}}{6!} −⋯+ \frac{(−1)^{k−1}x^{2k}}{(2k)!} + \frac{(−1)^{k}x^{2k+2}}{(2k + 2)!} +⋯+c_{1} +1 =

= − [1 − \frac{x^{2}}{2!} + \frac{x^{4}}{4!} − \frac{x^{6}}{6!} + ⋯ + ⋯ + \frac{(−1)^{k}x^{2k}}{(2k)!} + \frac{(−1)^{k+1}x^{2k+2}}{(2k + 2)!} + ⋯] + c = − cos x + c, for x∈R.

∫ \frac{sin x}{x} dx = ∫ [∑_{k=0}^{∞} \frac{(−1)^{k}x^{2k}}{(2k + 1)!}] dx = ∑_{k=0}^{∞} ∫ \frac{(−1)^{k}x^{2k}}{(2k + 1)!} dx = ∑_{k=0}^{∞} \frac{(−1)^{k}x^{2k+1}}{(2k + 1)(2k + 1)!} + c, for x∈R − \{0\}.

∫ \frac{sin x}{x} dx = ∫ [1 − \frac{x^{2}}{3!} + \frac{x^{4}}{5!} − \frac{x^{6}}{7!} + ⋯ + \frac{(−1)^{k}x^{2k}}{(2k + 1)!} + ⋯] dx = ∫ dx−∫ \frac{x^{2}}{3!} dx+∫ \frac{x^{4}}{5!} dx−∫ \frac{x^{6}}{7!} dx+⋯+∫ \frac{(−1)^{k}x^{2k}}{(2k + 1)!} dx+⋯ =

= x − \frac{x^{3}}{3 · 3!} + \frac{x^{5}}{5 · 5!} − \frac{x^{7}}{7 · 7!} + ⋯ + \frac{(−1)^{k}x^{2k+1}}{(2k + 1)(2k + 1)!} + ⋯ + c, for x∈R − \{0\}.

∫ cos x dx = ∫ [∑_{k=0}^{∞} \frac{(−1)^{k}x^{2k}}{(2k)!} ] dx = ∑_{k=0}^{∞} ∫ \frac{(−1)^{k}x^{2k}}{(2k)!} dx = ∑_{k=0}^{∞} \frac{(−1)^{k}x^{2k+1}}{(2k + 1)(2k)!} + c = ∑_{k=0}^{∞} \frac{(−1)^{k}x^{2k+1}}{(2k + 1)!} + c = sin x + c, for x∈R.

∫ cos x dx = ∫ [1 − \frac{x^{2}}{2!} + \frac{x^{4}}{4!} − \frac{x^{6}}{6!} + ⋯ + \frac{(−1)^{k}x^{2k}}{(2k)!} + ⋯] dx = ∫ dx−∫ \frac{x^{2}}{2!} dx+∫ \frac{x^{4}}{4!} dx−∫ \frac{x^{6}}{6!} dx+⋯+∫ \frac{(−1)^{k}x^{2k}}{(2k)!} dx+⋯ =

= x − \frac{x^{3}}{3 · 2!} + \frac{x^{5}}{5 · 4!} − \frac{x^{7}}{7 · 6!} + ⋯ + \frac{(−1)^{k}x^{2k+1}}{(2k + 1)(2k)!} + ⋯ + c = x − \frac{x^{3}}{3!} + \frac{x^{5}}{5!} − \frac{x^{7}}{7!} + ⋯ + \frac{(−1)^{k}x^{2k+1}}{(2k + 1)!} + ⋯ + c = sin x + c,

for x∈R.

∫ \frac{cos x}{x} dx = ∫ [∑_{k=0}^{∞} \frac{(−1)^{k}x^{2k−1}}{(2k)!} ] dx = ∫ [\frac{1}{x} + ∑_{k=1}^{∞} \frac{(−1)^{k}x^{2k−1}}{(2k)!} ] dx = ∫ \frac{dx}{x} + ∑_{k=1}^{∞} ∫ \frac{(−1)^{k}x^{2k−1}}{(2k)!} dx = ln |x| + ∑_{k=1}^{∞} \frac{(−1)^{k}x^{2k}}{2k(2k)!} ,

for x∈R−{0}.

∫ \frac{cos x}{x} dx = ∫ [\frac{1}{x} − \frac{x}{2!} + \frac{x^{3}}{4!} − \frac{x^{5}}{6!} + ⋯ + \frac{(−1)^{k}x^{2k−1}}{(2k)!} + ⋯] dx = ∫ \frac{dx}{x}−∫ \frac{x}{2!} dx+∫ \frac{x^{3}}{4!} dx−∫ \frac{x^{5}}{6!} dx+⋯+∫ \frac{(−1)^{k}x^{2k}}{2k(2k)!} dx+⋯ =

= ln |x| − \frac{x^{2}}{2 · 2!} + \frac{x^{4}}{4 · 4!} − \frac{x^{6}}{6 · 6!} + ⋯ + \frac{(−1)^{k}x^{2k}}{2k(2k)!} + ⋯ + c, for x∈R − \{0\}.

\underline{Integrals}

Examples

∫ [\sqrt{x^{3}} − \frac{1}{\sqrt{x}}] dx = ∫ [x^{\frac{3}{2}} − x^{− \frac{1}{2}} ] dx = \frac{2}{5}x^{\frac{5}{2}} − 2x^{\frac{1}{2}} + c = \frac{2\sqrt{x^{5}}}{5} − 2\sqrt{x} + c, for x > 0.

∫ x(x − a)(x − b) dx = ∫ [x^{3} − (a + b)x^{2} + abx] dx = \frac{1}{4}x^{4} − \frac{a + b}{3}x^{3} + \frac{ab}{2}x^{2} + c, for a, b∈R, x∈R.

∫ e^{ax} dx = \frac{1}{a} ∫ a e^{ax} dx = \begin{matrix}ax = t \\ a dx = dt \end{matrix} = \frac{1}{a} ∫ e^{t} dt = \frac{e^{t}}{a} + c = \frac{e^{ax}}{a} + c, for a∈R, a ≠ 0, x∈R.

∫ sinh^{2} x dx = ∫ \frac{(e^{x} − e^{−x})^{2}}{4} dx = \frac{1}{4} ∫ [e^{2x} −2 e^{x} e^{−x} + e^{−2x}] dx = \frac{1}{4} [\frac{e^{2x}}{2} − 2x + \frac{e^{−2x}}{−2} ] + c = \frac{e^{2x}}{8} − \frac{e^{−2x}}{8} − \frac{x}{2} + c, for x∈R.

∫ cosh^{2} x dx = ∫ \frac{(e^{x} + e^{−x})^{2}}{4} dx = \frac{1}{4} ∫ [e^{2x} +2 e^{x} e^{−x} + e^{−2x}] dx = \frac{1}{4} [\frac{e^{2x}}{2} + 2x + \frac{e^{−2x}}{−2} ] + c = \frac{e^{2x}}{8} − \frac{e^{−2x}}{8} + \frac{x}{2} + c, for x∈R.

∫ [tg x + cotg x] dx = ∫ [\frac{sin x}{cos x} + \frac{cos x}{sin x}] dx = ∫ [\frac{−(cos x)′}{cos x} + \frac{(sin x)′}{sin x} ] dx = − ln |cos x|+ln |sin x|+c = ln |\frac{sin x}{cos x}|+c = ln |tg x| + c,

for x∈R − \{k \frac{π}{2} ; k ∈Z\}.

∫ \frac{x dx}{x^{2} − a^{2}} = \frac{1}{2} ∫ \frac{2x dx}{x^{2} − a^{2}} = \frac{1}{2} ln |x^{2} − a^{2}| + c, for a∈R, x∈R − \{±a\}.

∫ \frac{x dx}{x^{2} + a^{2}} = \frac{1}{2} ∫ \frac{2x dx}{x^{2} + a^{2}} = \frac{1}{2} ln |x^{2} + a^{2}| + c = \frac{1}{2} ln (x^{2} + a^{2}) + c, for a∈R, a≠0, x∈R.

∫ \frac{dx}{x} = ∫ \frac{x dx}{x^{2} − 0} = \frac{1}{2} ln |x^{2}| + c = \frac{2}{2} ln |x| + c = ln |x| + c, for x∈R − \{0\}.

∫ \frac{x dx}{(x^{2} − a)^{n}} = \frac{1}{2} ∫ \frac{2x dx}{(x^{2} − a)^{n}} = \begin{matrix}x^{2} − a = t \\ 2x dx = dt \end{matrix} = \frac{1}{2} ∫ \frac{dt}{t^{n}} = \frac{1}{2} ∫ t^{−n} dt = \frac{t^{1−n}}{2(1 − n)} + c = \frac{(x^{2} − a)^{1−n}}{2(1 − n)} + c,

for n∈N, a ≥ 0, x∈R − \{±\sqrt{a}\}, a ≥ 0, resp. for a < 0, x∈R.

∫ \frac{ln cos x}{sin^{2} x} dx = u = ln cos x ⇒ u′ = \frac{− sin x}{cos x} v′ = \frac{1}{sin^{2} x} ⇒ v = − cotg x = −\frac{cos x}{sin x} = − cotg x ln cos x − ∫ dx = − cotg x ln cos x − x + c,

for x∈ ⋃_{k∈Z} [⟨−\frac{π}{2} + 2kπ; 2kπ) ∪ (2kπ; \frac{π}{2} + 2kπ⟩].

∫ \frac{dx}{x^{6} + 1} = ∫ [\frac{\frac{1}{3}}{x^{2} + 1} + \frac{−\frac{x}{2\sqrt{3}} + \frac{1}{3}}{x^{2} − \sqrt{3}x + 1} + \frac{\frac{x}{2\sqrt{3}} + \frac{1}{3}}{x^{2} + \sqrt{3}x + 1}] dx = \frac{1}{3} ∫ \frac{dx}{x^{2} + 1} + \frac{1}{2\sqrt{3}} ∫ [\frac{x + \frac{2\sqrt{3}}{3}}{x^{2} + \sqrt{3}x + 1} − \frac{x − \frac{2\sqrt{3}}{3}}{x^{2} − \sqrt{3}x + 1}] dx =

\frac{1}{x^{6} + 1} = \frac{Ax + B}{x^{2} + 1} + \frac{Cx + D}{x^{2} − \sqrt{3}x + 1} + \frac{Ex + F}{x^{2} + \sqrt{3}x + 1} ⟹ A = 0, B = \frac{1}{3}, C = −\frac{1}{2\sqrt{3}}, D = \frac{1}{3}, E = \frac{1}{2\sqrt{3}}, F = \frac{1}{3}

= \frac{1}{3} arctg x + \frac{1}{4\sqrt{3}} ∫ [\frac{2x + \frac{4\sqrt{3}}{3}}{x^{2} + \sqrt{3}x + 1} − \frac{2x − \frac{4\sqrt{3}}{3}}{x^{2} − \sqrt{3}x + 1}] dx =

= \frac{1}{3} arctg x + \frac{1}{4\sqrt{3}} ∫ [\frac{2x + \frac{4\sqrt{3}}{3} + \sqrt{3} − \sqrt{3}}{x^{2} + \sqrt{3}x + 1} − \frac{2x − \frac{4\sqrt{3}}{3} − \sqrt{3} + \sqrt{3}}{x^{2} − \sqrt{3}x + 1}] dx =

x^{2} ± \sqrt{3}x + 1 = (x ± \frac{\sqrt{3}}{2})^{2} + 1 − \frac{3}{4} = (x ± \frac{\sqrt{3}}{2})^{2} + \frac{1}{4} = (x ± \frac{\sqrt{3}}{2})^{2} + (\frac{1}{2})^{2} > 0

= \frac{1}{3} arctg x + \frac{1}{4\sqrt{3}} ∫ [\frac{2x + \sqrt{3}}{x^{2} + \sqrt{3}x + 1} − \frac{2x − \sqrt{3}}{x^{2} − \sqrt{3}x + 1}] dx + \frac{1}{4\sqrt{3}} ∫ [\frac{\frac{\sqrt{3}}{3}}{(x + \frac{\sqrt{3}}{2})^{2} + (\frac{1}{2})^{2}} − \frac{−\frac{\sqrt{3}}{3}}{(x − \frac{\sqrt{3}}{2})^{2} + (\frac{1}{2})^{2}} ] dx =

= \frac{1}{3} arctg x + \frac{1}{4\sqrt{3}} ∫ \frac{2x + \sqrt{3}}{x^{2} + \sqrt{3}x + 1} dx − \frac{1}{4\sqrt{3}} ∫ \frac{2x − \sqrt{3}}{x^{2} − \sqrt{3}x + 1} dx + \frac{1}{12} ∫ \frac{dx}{(x + \frac{\sqrt{3}}{2})^{2} + (\frac{1}{2})^{2}} + \frac{1}{12} ∫ \frac{dx}{(x − \frac{\sqrt{3}}{2})^{2} + (\frac{1}{2})^{2}} =

x + \frac{\sqrt{3}}{2} = t, dx = dt, x − \frac{\sqrt{3}}{2} = u, dx = du

= \frac{1}{3} arctg x + \frac{1}{4\sqrt{3}} ln |x^{2} + \sqrt{3}x + 1| − \frac{1}{4\sqrt{3}} ln |x^{2} − \sqrt{3}x + 1| + \frac{1}{12} ∫ \frac{dt}{t^{2} + (\frac{1}{2})^{2}} + \frac{1}{12} ∫ \frac{du}{u^{2} + (\frac{1}{2})^{2}} =

= \frac{1}{3} arctg x + \frac{1}{4\sqrt{3}} ln (x^{2} + \sqrt{3}x + 1) − \frac{1}{4\sqrt{3}} ln (x^{2} − \sqrt{3}x + 1) + \frac{2}{12} arctg 2t + \frac{2}{12} arctg 2u + c =

= \frac{1}{3} arctg x + \frac{1}{4\sqrt{3}} ln \frac{x^{2} + \sqrt{3}x + 1}{x^{2} − \sqrt{3}x + 1} + \frac{1}{6} arctg (2x + \sqrt{3}) + \frac{1}{6} arctg (2x − \sqrt{3}) + c, for x∈R.

∫ \frac{x dx}{x^{6} + 1} = \begin{matrix}x^{2} = t \\ 2x dx = dt \end{matrix} = \frac{1}{2} ∫ \frac{dt}{t^{3} + 1} = \frac{1}{2} ∫ \frac{dt}{(t + 1)(t^{2} − t + 1)} = \frac{1}{2} ∫ [\frac{\frac{1}{3}}{t + 1} + \frac{−\frac{t}{3} + \frac{2}{3}}{t^{2} − t + 1}] dt =

\frac{1}{t^{3} + 1} = \frac{A}{t + 1} + \frac{Bt + C}{t^{2} − t + 1} ⟹ 1 = A(t^{2} − t + 1) + (Bt + C)(t + 1) ⟹ A = \frac{1}{3}, B = −\frac{1}{3}, C = \frac{2}{3}

= \frac{1}{6} ∫ \frac{dt}{t + 1} − \frac{1}{6} ∫ \frac{t − 2}{t^{2} − t + 1} dt = \frac{1}{6} ∫ \frac{dt}{t + 1} − \frac{1}{12} ∫ \frac{2t − 4}{t^{2} − t + 1} dt = \frac{1}{6} ∫ \frac{dt}{t + 1} − \frac{1}{12} ∫ \frac{2t − 1 − 3}{t^{2} − t + 1} dt =

t^{2} − t + 1 = (t − \frac{1}{2})^{2} + 1 − \frac{1}{4} = (t − \frac{1}{2})^{2} + \frac{3}{4} = (t − \frac{1}{2})^{2} + (\frac{\sqrt{3}}{2})^{2} > 0

= \frac{1}{6} ln |t + 1| − \frac{1}{12} ∫ \frac{2t − 1}{t^{2} − t + 1} dt + \frac{1}{12} ∫ \frac{3 dt}{t^{2} − t + 1} = \frac{1}{12} ln (t + 1)^{2} − \frac{1}{12} ln |t^{2} − t + 1| + \frac{1}{4} ∫ \frac{dt}{(t − \frac{1}{2})^{2} + (\frac{\sqrt{3}}{2})^{2}} =

= t − \frac{1}{2} = z, dt = dz = \frac{1}{12} ln (t + 1)^{2} − \frac{1}{12} ln (t^{2} − t + 1) + \frac{1}{4} ∫ \frac{dz}{z^{2} + (\frac{\sqrt{3}}{2})^{2}} = \frac{1}{12} ln \frac{(t + 1)^{2}}{t^{2} − t + 1} + \frac{1}{4} \frac{2}{\sqrt{3}} arctg \frac{2z}{\sqrt{3}} + c_{1} =

= \frac{1}{12} ln \frac{t^{2} + 2t + 1}{t^{2} − t + 1} + \frac{1}{2\sqrt{3}} arctg \frac{2t − 1}{\sqrt{3}} + c_{1} = \frac{1}{12} ln \frac{x^{4} + 2x^{2} + 1}{x^{4} − x^{2} + 1} + \frac{\sqrt{3}}{6} arctg \frac{2x^{2} − 1}{\sqrt{3}} + c_{1}, for x∈R.

∫ \frac{x dx}{x^{6} + 1} = ∫ [\frac{\frac{x}{3}}{x^{2} + 1} + \frac{−\frac{x}{6} + \frac{\sqrt{3}}{6}}{x^{2} − \sqrt{3}x + 1} + \frac{−\frac{x}{6} − \frac{\sqrt{3}}{6}}{x^{2} + \sqrt{3}x + 1}] dx = \frac{1}{6} ∫ \frac{2x dx}{x^{2} + 1} − \frac{1}{12} ∫ [\frac{2x − 2\sqrt{3}}{x^{2} − \sqrt{3}x + 1} + \frac{2x + 2\sqrt{3}}{x^{2} + \sqrt{3}x + 1}] dx =

\frac{x}{x^{6} + 1} = \frac{Ax + B}{x^{2} + 1} + \frac{Cx + D}{x^{2} − \sqrt{3}x + 1} + \frac{Ex + F}{x^{2} + \sqrt{3}x + 1} ⟹ A = \frac{1}{3}, B = 0, C = −\frac{1}{6}, D = \frac{\sqrt{3}}{6}, E = −\frac{1}{6}, F = −\frac{\sqrt{3}}{6}

= \frac{1}{6} ln |x^{2} + 1| − \frac{1}{12} ∫ [\frac{2x − \sqrt{3} − \sqrt{3}}{x^{2} − \sqrt{3}x + 1} + \frac{2x + \sqrt{3} + \sqrt{3}}{x^{2} + \sqrt{3}x + 1}] dx = x^{2} ± \sqrt{3}x + 1 = (x ± \frac{\sqrt{3}}{2})^{2} + (\frac{1}{2})^{2} > 0 =

= \frac{1}{12} ln (x^{2} + 1)^{2} − \frac{1}{12} ∫ [\frac{2x − \sqrt{3}}{x^{2} − \sqrt{3}x + 1} + \frac{2x + \sqrt{3}}{x^{2} + \sqrt{3}x + 1}] dx + \frac{1}{12} ∫ [\frac{\sqrt{3}}{(x − \frac{\sqrt{3}}{2})^{2} + (\frac{1}{2})^{2}} − \frac{\sqrt{3}}{(x + \frac{\sqrt{3}}{2})^{2} + (\frac{1}{2})^{2}} ] dx =

x − \frac{\sqrt{3}}{2} = t, dx = dt,  x + \frac{\sqrt{3}}{2} = u, dx = du  (x^{2} − \sqrt{3}x + 1) (x^{2} + \sqrt{3}x + 1) = x^{4} − x^{2} + 1

= \frac{1}{12} ln (x^{4} + 2x^{2} + 1) − \frac{1}{12} ln |x^{2} − \sqrt{3}x + 1| − \frac{1}{12} ln |x^{2} + \sqrt{3}x + 1| + \frac{\sqrt{3}}{12} ∫ \frac{dt}{t^{2} + (\frac{1}{2})^{2}} − \frac{\sqrt{3}}{12} ∫ \frac{du}{u^{2} + (\frac{1}{2})^{2}} =

= \frac{1}{12} ln (x^{4} + 2x^{2} + 1) − \frac{1}{12} ln (x^{2} − \sqrt{3}x + 1) − \frac{1}{12} ln (x^{2} + \sqrt{3}x + 1) + \frac{2\sqrt{3}}{12} arctg 2t − \frac{2\sqrt{3}}{12} arctg 2u + c_{2} =

= \frac{1}{12} ln \frac{x^{4} + 2x^{2} + 1}{x^{4} − x^{2} + 1} + \frac{\sqrt{3}}{6} arctg (2x − \sqrt{3}) − \frac{\sqrt{3}}{6} arctg (2x + \sqrt{3}) + c_{2}, for x∈R.

?? \frac{1}{12} ln \frac{x^{4} + 2x^{2} + 1}{x^{4} − x^{2} + 1} + \frac{\sqrt{3}}{6} arctg \frac{2x^{2} − 1}{\sqrt{3}} + c_{1} {? \atop =} \frac{1}{12} ln \frac{x^{4} + 2x^{2} + 1}{x^{4} − x^{2} + 1} + \frac{\sqrt{3}}{6} [arctg (2x − \sqrt{3}) − arctg (2x + \sqrt{3})] + c_{2} ??

[\frac{1}{12} ln \frac{x^{4} + 2x^{2} + 1}{x^{4} − x^{2} + 1} + \frac{\sqrt{3}}{6} arctg \frac{2x^{2} − 1}{\sqrt{3}} + c_{1}] − [\frac{1}{12} ln \frac{x^{4} + 2x^{2} + 1}{x^{4} − x^{2} + 1} + \frac{\sqrt{3}}{6} arctg (2x − \sqrt{3}) − \frac{\sqrt{3}}{6} arctg (2x + \sqrt{3}) + c_{2}] =

= c_{1} − c_{2} + \frac{\sqrt{3}}{6} [arctg \frac{2x^{2} − 1}{\sqrt{3}} − arctg (2x − \sqrt{3}) + arctg (2x + \sqrt{3})] = arctg α = \frac{π}{2} − arctg α for α∈R =

= c_{1} − c_{2} + \frac{\sqrt{3}}{6} [\frac{π}{2} − arccotg \frac{2x^{2} − 1}{\sqrt{3}} − \frac{π}{2} + arccotg (2x − \sqrt{3}) + \frac{π}{2} − arccotg (2x + \sqrt{3})] = c_{1} − c_{2} + \frac{π\sqrt{3}}{2 · 6} = c =

= c+ \frac{\sqrt{3}}{6} [arccotg (2x−\sqrt{3})− arccotg (2x+\sqrt{3})− arccotg \frac{2x^{2}−1}{\sqrt{3}}] = arccotg α − arccotg β = arccotg \frac{αβ+1}{β − α} for α, β ∈R, α ≠ β =

= c + \frac{\sqrt{3}}{6} [arccotg \frac{(2x − \sqrt{3}) (2x + \sqrt{3}) + 1}{(2x + \sqrt{3}) − (2x − \sqrt{3})} − arccotg \frac{2x^{2} − 1}{\sqrt{3}}] = c + \frac{\sqrt{3}}{6} [arccotg \frac{4x^{2} − 3 + 1}{2\sqrt{3}} − arccotg \frac{2x^{2} − 1}{\sqrt{3}}] =

= c + \frac{\sqrt{3}}{6} [arccotg \frac{4x^{2} − 2}{2\sqrt{3}} − arccotg \frac{2x^{2} − 1}{\sqrt{3}}] = c + \frac{\sqrt{3}}{6} [arccotg \frac{2x^{2} − 1}{\sqrt{3}} − arccotg \frac{2x^{2} − 1}{\sqrt{3}}] = c = c_{1} − c_{2} + \frac{π\sqrt{3}}{12} (i.e. const.).

⟹ ∫ \frac{x dx}{x^{6} + 1} = \frac{1}{12} ln \frac{x^{4} + 2x^{2} + 1}{x^{4} − x^{2} + 1} + \frac{\sqrt{3}}{6} arctg \frac{2x^{2} − 1}{\sqrt{3}}+ c_{1} =

= \frac{1}{12} ln \frac{x^{4} + 2x^{2} + 1}{x^{4} − x^{2} + 1} + \frac{\sqrt{3}}{6} [arctg (2x − \sqrt{3}) − arctg (2x + \sqrt{3})]+ c_{1} − \frac{π\sqrt{3}}{12}, for x∈R.

∫ \frac{x^{2} dx}{x^{6} + 1} = \frac{1}{3} ∫ \frac{3x^{2} dx}{x^{6} + 1} = \begin{matrix}x^{3} = t \\ 3x^{2} dx = dt \end{matrix} = \frac{1}{3} ∫ \frac{dt}{t^{2} + 1} = \frac{1}{3} arctg t + c_{1} = \frac{1}{3} arctg x^{3} + c_{1}, for x∈R.

∫ \frac{x^{2} dx}{x^{6} + 1} = ∫ [\frac{−\frac{1}{3}}{x^{2} + 1} + \frac{\frac{1}{6}}{x^{2} − \sqrt{3}x + 1} + \frac{\frac{1}{6}}{x^{2} + \sqrt{3}x + 1}] dx = −\frac{1}{3} ∫ \frac{dx}{x^{2} + 1} + \frac{1}{6} ∫ \frac{dx}{x^{2} − \sqrt{3}x + 1} + \frac{1}{6} ∫ \frac{dx}{x^{2} + \sqrt{3}x + 1} =

\frac{x^{2}}{x^{6} + 1} = \frac{Ax + B}{x^{2} + 1} + \frac{Cx + D}{x^{2} − \sqrt{3}x + 1} + \frac{Ex + F}{x^{2} + \sqrt{3}x + 1} ⟹ A = 0, B = −\frac{1}{3}, C = 0, D = \frac{1}{6}, E = 0, F = \frac{1}{6}

= x^{2} ± \sqrt{3}x + 1 = (x ± \frac{\sqrt{3}}{2})^{2} + (\frac{1}{2})^{2} > 0 = \frac{1}{6} ∫ \frac{dx}{(x − \frac{\sqrt{3}}{2})^{2} + (\frac{1}{2})^{2}} + \frac{1}{6} ∫ \frac{dx}{(x + \frac{\sqrt{3}}{2})^{2} + (\frac{1}{2})^{2}} − \frac{1}{3} arctg x =

= x − \frac{\sqrt{3}}{2} = t, dx = dt, x + \frac{\sqrt{3}}{2} = u, dx = du = \frac{1}{6} ∫ \frac{dt}{t^{2} + (\frac{1}{2})^{2}} + \frac{1}{6} ∫ \frac{du}{u^{2} + (\frac{1}{2})^{2}} − \frac{1}{3} arctg x =

= \frac{2}{6} arctg 2t + \frac{2}{6} arctg 2u − \frac{1}{3} arctg x + c_{2} = \frac{1}{3} arctg (2x − \sqrt{3}) + \frac{1}{3} arctg (2x + \sqrt{3}) − \frac{1}{3} arctg x + c_{2}, for x∈R.

?? \frac{1}{3} arctg x^{3} + c_{1} {? \atop =} \frac{1}{3} arctg (2x − \sqrt{3}) + \frac{1}{3} arctg (2x + \sqrt{3}) − \frac{1}{3} arctg x + c_{2} ??

E(x) = [\frac{1}{3} arctg x^{3} + c_{1}] − [\frac{1}{3} arctg (2x − \sqrt{3}) + \frac{1}{3} arctg (2x + \sqrt{3}) − \frac{1}{3} arctg x + c_{2}] = arctg α = \frac{π}{2} − arctg α for α∈R =

= \frac{1}{3} [\frac{π}{2} − arccotg x^{3} − \frac{π}{2} + arccotg (2x − \sqrt{3}) − \frac{π}{2} + arccotg (2x + \sqrt{3}) + \frac{π}{2} − arccotg x] + c_{1} − c_{2} =

= \frac{1}{3} [arccotg (2x − \sqrt{3}) + arccotg (2x + \sqrt{3}) − arccotg x − arccotg x^{3}] + c_{1} − c_{2} =

arccotg α + arccotg β = arccotg \frac{αβ − 1}{α + β} for α, β ∈R, α ≠ −β 2x − \sqrt{3} ≠ − (2x + \sqrt{3}) ⟹ x ≠ \frac{\sqrt{3}}{2} x ≠ −x^{3} ⟹ x ≠ 0

= \frac{1}{3} [arccotg \frac{(2x−\sqrt{3}) (2x+\sqrt{3})−1}{(2x−\sqrt{3}) + (2x+\sqrt{3})} − arccotg \frac{x · x^{3} − 1}{x + x^{3}}] + c_{1} − c_{2} = \frac{1}{3} [arccotg \frac{4x^{2} − 3 − 1}{4x} − arccotg \frac{x^{4} − 1}{x + x^{3}}] + c_{1} − c_{2} =

= \frac{1}{3} [arccotg \frac{x^{2} − 1}{x} − arccotg \frac{(x^{2} − 1)(x^{2} + 1)}{x(1 + x^{2})}]+c_{1}−c_{2} = \frac{1}{3} [arccotg \frac{x^{2} − 1}{x} − arccotg \frac{x^{2} − 1}{x}]+c_{1}−c_{2} = c_{1} − c_{2} (i.e. const.).

for x∈R − \{0, \frac{\sqrt{3}}{2}\}.

E(0) = [\frac{1}{3} arctg 0^{3} + c_{1}] − [\frac{1}{3} arctg (2 · 0 − \sqrt{3}) + \frac{1}{3} arctg (2 · 0 + \sqrt{3}) − \frac{1}{3} arctg 0 + c_{2}] =

= \frac{1}{3} arctg 0 + c_{1} − \frac{1}{3} arctg (−\sqrt{3}) − \frac{1}{3} arctg \sqrt{3} + \frac{1}{3} arctg 0 − c_{2} = c_{1} − c_{2}, for x = 0.

E (\frac{\sqrt{3}}{2}) = [\frac{1}{3} arctg (\frac{\sqrt{3}}{2})^{3} + c_{1}] − [\frac{1}{3} arctg (2\frac{\sqrt{3}}{2} − \sqrt{3}) + \frac{1}{3} arctg (2\frac{\sqrt{3}}{2} + \sqrt{3}) − \frac{1}{3} arctg \frac{\sqrt{3}}{2} + c_{2}] =

= \frac{1}{3} arctg \frac{3\sqrt{3}}{8} + c_{1} − \frac{1}{3} arctg 0 − \frac{1}{3} arctg 2\sqrt{3} + \frac{1}{3} arctg \frac{\sqrt{3}}{2} − c_{2} = c_{1} − c_{2} + \frac{1}{3} [arctg \frac{3\sqrt{3}}{8} + arctg \frac{\sqrt{3}}{2} − arctg 2\sqrt{3}] =

= arctg α − arctg β = arctg \frac{α − β}{1 + αβ} for α, β ∈R, αβ > −1 = c_{1} − c_{2} + \frac{1}{3} [arctg \frac{3\sqrt{3}}{8} + arctg \frac{\frac{\sqrt{3}}{2} − 2\sqrt{3}}{1 + \frac{\sqrt{3}}{2}2\sqrt{3}}] =

= c_{1} − c_{2} + \frac{1}{3} [arctg \frac{3\sqrt{3}}{8} + arctg \frac{−\frac{3\sqrt{3}}{2}}{1 + 3} ] = c_{1} − c_{2} + \frac{1}{3} [arctg \frac{3\sqrt{3}}{8} + arctg \frac{−3\sqrt{3}}{8}] = c_{1} − c_{2}, for x = \frac{\sqrt{3}}{2}.

⟹ ∫ \frac{x^{2} dx}{x^{6} + 1} = \frac{1}{3} arctg x^{3}+c_{1} =

= \frac{1}{3} arctg (2x − \sqrt{3}) + \frac{1}{3} arctg (2x + \sqrt{3}) − \frac{1}{3} arctg x+c_{1}, for x∈R.

∫ \frac{x^{3} dx}{x^{6} + 1} = \frac{1}{2} ∫ \frac{2x · x^{2}}{x^{6} + 1} dx = \begin{matrix}x^{2} = t \\ 2x dx = dt \end{matrix} = \frac{1}{2} ∫ \frac{t dt}{t^{3} + 1} = \frac{1}{2} ∫ \frac{t dt}{(t + 1)(t^{2} − t + 1)} = \frac{1}{2} ∫ [\frac{−\frac{1}{3}}{t + 1} + \frac{\frac{t}{3} + \frac{1}{3}}{t^{2} − t + 1}] dt =

\frac{t}{t^{3} + 1} = \frac{A}{t + 1} + \frac{Bt + C}{t^{2} − t + 1} ⟹ t = A(t^{2} − t + 1) + (Bt + C)(t + 1) ⟹ A = −\frac{1}{3}, B = \frac{1}{3}, C = \frac{1}{3}

= \frac{1}{6} ∫ \frac{t + 1}{t^{2} − t + 1} dt − \frac{1}{6} ∫ \frac{dt}{t + 1} = \frac{1}{12} ∫ \frac{2t + 2}{t^{2} − t + 1} dt − \frac{1}{6} ∫ \frac{dt}{t + 1} = \frac{1}{12} ∫ \frac{2t − 1 + 3}{t^{2} − t + 1} dt − \frac{1}{6} ∫ \frac{dt}{t + 1} =

t^{2} − t + 1 = (t − \frac{1}{2})^{2} + 1 − \frac{1}{4} = (t − \frac{1}{2})^{2} + \frac{3}{4} = (t − \frac{1}{2})^{2} + (\frac{\sqrt{3}}{2})^{2} > 0

= \frac{1}{12} ∫ \frac{2t − 1}{t^{2} − t + 1} dt + \frac{1}{4} ∫ \frac{dt}{(t − \frac{1}{2})^{2} + (\frac{\sqrt{3}}{2})^{2}} − \frac{1}{6} ln |t + 1| = t − \frac{1}{2} = z, dt = dz =

= \frac{1}{12} ln (t^{2} − t + 1) + \frac{1}{4} ∫ \frac{dz}{z^{2} + (\frac{\sqrt{3}}{2})^{2}} − \frac{1}{12} ln (t + 1)^{2} = \frac{1}{12} ln \frac{t^{2} − t + 1}{(t + 1)^{2}} + \frac{1}{4} \frac{2}{\sqrt{3}} arctg \frac{2z}{\sqrt{3}} + c =

= \frac{1}{12} ln \frac{t^{2} − t + 1}{t^{2} + 2t + 1} + \frac{1}{2\sqrt{3}} arctg \frac{2t − 1}{\sqrt{3}} + c = \frac{1}{12} ln \frac{x^{4} − x^{2} + 1}{x^{4} + 2x^{2} + 1} + \frac{1}{2\sqrt{3}} arctg \frac{2x^{2} − 1}{\sqrt{3}} + c, for x∈R.

∫ \frac{x^{4} dx}{x^{6} + 1} = ∫ [\frac{\frac{1}{3}}{x^{2} + 1} + \frac{\frac{x}{2\sqrt{3}} − \frac{1}{6}}{x^{2} − \sqrt{3}x + 1} + \frac{−\frac{x}{2\sqrt{3}} − \frac{1}{6}}{x^{2} + \sqrt{3}x + 1}] dx = \frac{1}{3} ∫ \frac{dx}{x^{2} + 1} + \frac{1}{2\sqrt{3}} ∫ [\frac{x − \frac{\sqrt{3}}{3}}{x^{2} − \sqrt{3}x + 1} − \frac{x + \frac{\sqrt{3}}{3}}{x^{2} + \sqrt{3}x + 1}] dx =

\frac{x^{4}}{x^{6} + 1} = \frac{Ax + B}{x^{2} + 1} + \frac{Cx + D}{x^{2} − \sqrt{3}x + 1} + \frac{Ex + F}{x^{2} + \sqrt{3}x + 1} ⟹ A = 0, B = \frac{1}{3}, C = \frac{1}{2\sqrt{3}}, D = −\frac{1}{6}, E = −\frac{1}{2\sqrt{3}}, F = −\frac{1}{6}

= \frac{1}{3} arctg x + \frac{1}{4\sqrt{3}} ∫ [\frac{2x − \frac{2\sqrt{3}}{3} + \sqrt{3} − \sqrt{3}}{x^{2} − \sqrt{3}x + 1} − \frac{2x + \frac{2\sqrt{3}}{3} − \sqrt{3} + \sqrt{3}}{x^{2} + \sqrt{3}x + 1}] dx =

x^{2} ± \sqrt{3}x + 1 = (x ± \frac{\sqrt{3}}{2})^{2} + 1 − \frac{3}{4} = (x ± \frac{\sqrt{3}}{2})^{2} + (\frac{1}{2})^{2} > 0  x − \frac{\sqrt{3}}{2} = t, dx = dt  x + \frac{\sqrt{3}}{2} = u, dx = du

= \frac{arctg x}{3} + \frac{1}{4\sqrt{3}} ∫ [\frac{2x − \sqrt{3}}{x^{2} − \sqrt{3}x + 1} − \frac{2x + \sqrt{3}}{x^{2} + \sqrt{3}x + 1}] dx + \frac{1}{4\sqrt{3}} ∫ [\frac{\frac{\sqrt{3}}{3}}{(x − \frac{\sqrt{3}}{2})^{2} + (\frac{1}{2})^{2}} − \frac{−\frac{\sqrt{3}}{3}}{(x + \frac{\sqrt{3}}{2})^{2} + (\frac{1}{2})^{2}} ] dx =

= \frac{1}{3} arctg x + \frac{1}{4\sqrt{3}} ln |x^{2} − \sqrt{3}x + 1| − \frac{1}{4\sqrt{3}} ln |x^{2} + \sqrt{3}x + 1| + \frac{1}{12} ∫ \frac{dt}{t^{2} + (\frac{1}{2})^{2}} + \frac{1}{12} ∫ \frac{du}{u^{2} + (\frac{1}{2})^{2}} =

= \frac{1}{3} arctg x + \frac{1}{4\sqrt{3}} ln (x^{2} − \sqrt{3}x + 1) − \frac{1}{4\sqrt{3}} ln (x^{2} + \sqrt{3}x + 1) + \frac{2}{12} arctg 2t + \frac{2}{12} arctg 2u + c =

= \frac{1}{3} arctg x + \frac{1}{4\sqrt{3}} ln \frac{x^{2} − \sqrt{3}x + 1}{x^{2} + \sqrt{3}x + 1} + \frac{1}{6} arctg (2x − \sqrt{3}) + \frac{1}{6} arctg (2x + \sqrt{3}) + c, for x∈R.

∫ \frac{x^{5} dx}{x^{6} + 1} = \frac{1}{6} ∫ \frac{6x^{5}}{x^{6} + 1} dx = \begin{matrix}x^{6} = t \\ 6x^{5} dx = dt \end{matrix} = \frac{1}{3} ∫ \frac{dt}{t + 1} = \frac{1}{6} ln |t + 1| + c = \frac{1}{6} ln |x^{6} + 1| + c = \frac{1}{6} ln (x^{6} + 1) + c, for x∈R.

∫ \frac{x^{6} dx}{x^{6} + 1} = ∫ \frac{x^{6} + 1 − 1}{x^{6} + 1} dx = ∫ [1 − \frac{1}{x^{6} + 1}] dx = ∫ dx − ∫ \frac{dx}{x^{6} + 1} =

= x − \frac{1}{3} arctg x − \frac{1}{4\sqrt{3}} ln \frac{x^{2} + \sqrt{3}x + 1}{x^{2} − \sqrt{3}x + 1} − \frac{1}{6} arctg (2x + \sqrt{3}) − \frac{1}{6} arctg (2x − \sqrt{3}) + c, for x∈R.

∫ \frac{dx}{x ln x} = ∫ \frac{\frac{1}{x}}{ln x} dx = ∫ \frac{[ln x]′}{ln x} dx = ln |ln x| + c, for x > 0.

∫ x^{2} ln x dx = \begin{matrix}u = ln x ⇒ & u′ = \frac{1}{x} \\ v′ = x^{2} ⇒ & v = \frac{x^{3}}{3} \end{matrix} = \frac{x^{3}}{3} ln x − ∫ \frac{x^{3}}{3} \frac{1}{x} dx = \frac{x^{3}}{3} ln x − ∫ \frac{x^{2}}{3} dx = \frac{x^{3}}{3} ln x − \frac{x^{3}}{9} + c, for x > 0.

∫ \frac{ln x}{\sqrt{x}} dx = \begin{matrix}u = ln x ⇒ & u′ = \frac{1}{x} \\ v′ = x^{− \frac{1}{2}} ⇒ & v = 2x^{\frac{1}{2}} \end{matrix} = 2\sqrt{x} ln x−2 ∫ \frac{x^{\frac{1}{2}}}{x} dx = 2\sqrt{x} ln x−2 ∫ x^{− \frac{1}{2}} dx = 2\sqrt{x} ln x−2·2x^{\frac{1}{2}} +c = 2\sqrt{x} ln x − 4\sqrt{x} + c,

for x > 0.

∫ \frac{dx}{\sqrt{x(1 − x)}} = ∫ \frac{t^{2} + 1}{t} \frac{2t dt}{(t^{2} + 1)^{2}} = ∫ \frac{2 dt}{t^{2} + 1} = 2 arctg t + c_{1} = 2 arctg \sqrt{\frac{x}{1 − x}} + c_{1}, for x∈(0 ; 1).

x(1 − x) = x − x^{2} > 0 ⟹ x > x^{2} ⟹ x∈(0 ; 1)

3-rd Euler: t = \sqrt{\frac{x}{1 − x}} ⟹ t^{2} = \frac{x}{1 − x} ⟹ t^{2} − t^{2}x = x ⟹ x = \frac{t^{2}}{t^{2} + 1} ⟹ dx = \frac{2t(t^{2} + 1) − t^{2} · 2t}{(t^{2} + 1)^{2}} dt = \frac{2t dt}{(t^{2} + 1)^{2}}

x∈(0 ; 1) ⟹ t∈(0 ; ∞) , x(1 − x) = \frac{t^{2}}{t^{2} + 1} (1 − \frac{t^{2}}{t^{2} + 1}) = \frac{t^{2}(t^{2} + 1 − t^{2})}{(t^{2} + 1)^{2}} = \frac{t^{2}}{(t^{2} + 1)^{2}} ⟹ \sqrt{x(1 − x)} = \frac{t}{t^{2} + 1}

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∫ \frac{dx}{\sqrt{x(1 − x)}} = x(1 − x) = −[x^{2} − x] = − [(x − \frac{1}{2})^{2} − \frac{1}{4}] ⟹ x − \frac{1}{2} = t ⟹ dx = dt = ∫ \frac{dt}{\sqrt{(\frac{1}{2})^{2} − t^{2}}} = arcsin 2t + c_{2} =

= arcsin (2x − 1) + c_{2}, for x∈(0 ; 1).

∫ \frac{dx}{(x − \sqrt{x^{2} − 1})^{2}} = ∫ \frac{1}{(x − \sqrt{x^{2} − 1})^{2}} \begin{matrix}(x + \sqrt{x^{2} − 1}) & 2 \\ (x + \sqrt{x^{2} − 1}) & 2 \end{matrix} dx = ∫ \frac{x^{2} + 2x\sqrt{x^{2} − 1} + [x^{2} − 1]}{(x^{2} − [x^{2} − 1])^{2}} dx = ∫ \frac{2x^{2} − 1 + 2x\sqrt{x^{2} − 1}}{1^{2}} dx =

= ∫ (2x^{2} − 1) dx + ∫ 2x\sqrt{x^{2} − 1} dx = x^{2} − 1 = t ⇒ 2x dx = dt = ∫ (2x^{2} − 1) dx + ∫ t^{\frac{1}{2}} dt =

= \frac{2}{3}x^{3} − x + \frac{2}{3}t^{\frac{3}{2}} + c = \frac{2}{3}x^{3} − x + \frac{2}{3} (x^{2} − 1)^{\frac{3}{2}} + c = \frac{2}{3}x^{3} − x + \frac{2}{3}\sqrt{(x^{2} − 1)^{3}} + c, for x∈R − (−1 ; 1).

∫ \frac{1 + x}{\sqrt{1 − x^{2}}} dx = ∫ \frac{dx}{\sqrt{1 − x^{2}}} + ∫ \frac{x dx}{\sqrt{1 − x^{2}}} = arcsin x − \frac{1}{2} ∫ \frac{dt}{\sqrt{t}} = \begin{matrix}t = 1 − x^{2} \\ dt = −2x dx ⇒ x dx = − dt/2 \end{matrix} = arcsin x − \frac{1}{2} ∫ t^{− \frac{1}{2}} dt =

= arcsin x − \frac{1}{2} \frac{t^{\frac{1}{2}}}{\frac{1}{2}} + c = arcsin x − \sqrt{t} + c = arcsin x − \sqrt{1 − x^{2}} + c, for x∈(0 ; 1).

∫ \frac{1 + x}{\sqrt{1 − x^{2}}} dx = ∫ \frac{1 + x}{\sqrt{(1 − x)(1 + x)}} dx = ∫ \sqrt{\frac{1 + x}{1 − x}} dx = ∫ \frac{4t^{2} dt}{(t^{2} + 1)^{2}} = ∫ \frac{2t · 2t dt}{(t^{2} + 1)^{2}} = \begin{matrix}u = 2t ⇒ & u′ = 2 \\ v′ = \frac{2t}{(t^{2} + 1)^{2}} ⇒ & v = −\frac{1}{t^{2} + 1} \end{matrix} =

\sqrt{\frac{1 + x}{1 − x}} = t ⇒ \frac{1 + x}{1 − x} = t^{2} ⇒ 1 + x = t^{2} − t^{2}x ⇒ x = \frac{t^{1} − 1}{t^{2} + 1} ⇒ dx = \frac{2t(t^{2} + 1) − (t^{2} − 1)2t}{(t^{2} + 1)^{2}} dt = \frac{4t dt}{(t^{2} + 1)^{2}}

= −\frac{2t}{t^{2} + 1} − ∫ \frac{−2 dt}{t^{2} + 1} = 2 ∫ \frac{dt}{t^{2} + 1} − \frac{2t}{t^{2} + 1} = 2 arctg t − \frac{2t}{t^{2} + 1} + c = 2 arctg \sqrt{\frac{1 + x}{1 − x}} − \frac{2\sqrt{\frac{1 + x}{1 − x}}}{\frac{1 + x}{1 − x} + 1} + c =

= 2 arctg \sqrt{\frac{1 + x}{1 − x}} − \frac{2\sqrt{\frac{1 + x}{1 − x}}}{\frac{2}{1 − x}} +c = 2 arctg \sqrt{\frac{1 + x}{1 − x}} −\sqrt{1 − x^{2}}+ c = 2 arctg \sqrt{\frac{1 + x}{1 − x}} − \sqrt{1 − x^{2}} + c, for x∈(0 ; 1).

∫ \frac{x^{n}}{x − 1} dx = ∫ \frac{x^{n} − 1 + 1}{x − 1} dx = ∫ \frac{x^{n} − 1}{x − 1} dx + ∫ \frac{dx}{x − 1} = ∫ [x^{n−1} + x^{n−2} + ⋯ + x + 1] dx + ∫ \frac{dx}{x − 1} =

= \frac{x^{n}}{n} + \frac{x^{n−1}}{n − 1} + ⋯ + \frac{x^{2}}{2} + x + ln |x − 1| + c = ln |x − 1| + ∑_{k=1}^{n} \frac{x^{k}}{k} + c, for x∈R − \{1\}, n∈N.

∫ \frac{x}{x − 1} dx = ∫ \frac{x − 1 + 1}{x − 1} dx = ∫ dx + ∫ \frac{dx}{x − 1} = x + ln |x − 1| + c, for x∈R − \{1\}.

∫ \frac{x^{3}}{x − 1} dx = ∫ \frac{x^{3} − 1 + 1}{x − 1} dx = ∫ [x^{2} + x + 1] dx + ∫ \frac{dx}{x − 1} = \frac{x^{3}}{3} + \frac{x^{2}}{2} + x + ln |x − 1| + c, for x∈R − \{1\}.

∫ \frac{x^{9}}{x − 1} dx = \frac{x^{9}}{9} + \frac{x^{8}}{8} + \frac{x^{7}}{7} + \frac{x^{6}}{6} + \frac{x^{5}}{5} + \frac{x^{4}}{4} + \frac{x^{3}}{3} + \frac{x^{2}}{2} + x + ln |x − 1| + c = ln |x − 1| + ∑_{k=1}^{9} \frac{x^{k}}{k} + c, for x∈R − \{1\}.

∫ \frac{dx}{cos x} = t = tg \frac{x}{2}, dx = \frac{2 dt}{1 + t^{2}} , cos x = \frac{1 − t^{2}}{1 + t^{2}} = ∫ \frac{1 + t^{2}}{1 − t^{2}} \frac{2 dt}{1 + t^{2}} = ∫ \frac{2 dt}{1 − t^{2}} = − ∫ \frac{2 dt}{t^{2} − 1} = −\frac{2}{2} ln |\frac{t − 1}{t + 1}| + c =

= − ln |\frac{t − 1}{t + 1}| + c = ln |\frac{t + 1}{t − 1}| + c = ln |\frac{tg \frac{x}{2} + 1}{tg \frac{x}{2} − 1}| + c, for x∈R − \{\frac{π}{2} + kπ; k ∈Z\}.

∫ \frac{dx}{cos x} = ∫ \frac{cos x dx}{cos^{2} x} = ∫ \frac{cos x dx}{1 − sin^{2} x} = t = sin x ; dt = cos x dx = ∫ \frac{dt}{1 − t^{2}} = − ∫ \frac{dt}{t^{2} − 1} = −\frac{1}{2} ln |\frac{t − 1}{t + 1}| + c = \frac{1}{2} ln |\frac{t + 1}{t − 1}| + c =

= \frac{1}{2} ln |\frac{sin x + 1}{sin x − 1}| + c = \frac{1}{2} ln \frac{1 + sin x}{1 − sin x} + c = ln \sqrt{\frac{1 + sin x}{1 − sin x}} + c, for x∈R − \{\frac{π}{2} + kπ; k ∈Z\}.

∫ \frac{dx}{1 + cos x} = t = tg \frac{x}{2}, dx = \frac{2 dt}{1 + t^{2}} , 1 + cos x = 1 + \frac{1 − t^{2}}{1 + t^{2}} = \frac{2}{1 + t^{2}} = ∫ \frac{1 + t^{2}}{2} \frac{2 dt}{1 + t^{2}} = ∫ dt = t + c = tg \frac{x}{2} + c,

for x∈R − {π + 2kπ; k ∈Z}.

∫ \frac{dx}{1 + cos x} = ∫ \frac{1 − cos x}{(1 + cos x)(1 − cos x)} dx = ∫ \frac{1 − cos x}{1 − cos^{2} x} dx = ∫ \frac{dx}{sin^{2} x} − ∫ \frac{cos x}{sin^{2} x} dx = sin x = t ; cos x dx = dt = − cotg x − ∫ \frac{dt}{t^{2}} =

= − cotg x− \frac{t^{−1}}{−1} +c = − cotg x+ \frac{1}{t} +c = − cotg x+ \frac{1}{sin x} +c = −\frac{cos x}{sin x} + \frac{1}{sin x} +c = \frac{1 − cos x}{sin x} + c, for x∈R−\{kπ; k ∈Z\}.

∫ \frac{dx}{sin x} = t = tg \frac{x}{2}, dx = \frac{2 dt}{1 + t^{2}} , sin x = \frac{2t}{1 + t^{2}} = ∫ \frac{1 + t^{2}}{2t} \frac{2 dt}{1 + t^{2}} = ∫ \frac{dt}{t} = ln |t|+c = ln |tg \frac{x}{2}| + c, for x∈R−\{kπ; k ∈Z\}.

∫ \frac{dx}{sin x} = ∫ \frac{sin x dx}{sin^{2} x} = ∫ \frac{sin x dx}{1 − cos^{2} x} = t = cos x dt = − sin x dx = − ∫ \frac{dt}{1 − t^{2}} = ∫ \frac{dt}{t^{2} − 1} = \frac{1}{2} ln |\frac{t − 1}{t + 1}| + c = \frac{1}{2} ln |\frac{cos x − 1}{cos x + 1}| + c =

= \frac{1}{2} ln \frac{1 − cos x}{1 + cos x} + c = ln \sqrt{\frac{1 − cos x}{1 + cos x}} + c, for x∈R − \{kπ; k ∈Z\}.

∫ \frac{dx}{1 + sin x} = t = tg \frac{x}{2}, dx = \frac{2 dt}{1 + t^{2}} , 1 + sin x = 1 + \frac{2t}{1 + t^{2}} = \frac{1 + 2t + t^{2}}{1 + t^{2}} = ∫ \frac{1 + t^{2}}{(1 + t)^{2}} \frac{2 dt}{1 + t^{2}} = ∫ \frac{2 dt}{(1 + t)^{2}} = \begin{matrix}1 + t = u \\ dt = du \end{matrix} =

= 2 ∫ \frac{du}{u^{2}} = 2 ∫ u^{−2} du = 2\frac{u^{−1}}{−1} + c = c − \frac{2}{u} = c − \frac{2}{t + 1} = c − \frac{2}{tg \frac{x}{2} + 1}, for x∈R − \{\frac{π}{2} + 2kπ; k ∈Z\}.

∫ \frac{dx}{1 + sin x} = ∫ \frac{1 − sin x}{(1 + sin x)(1 − sin x)} dx = ∫ \frac{1 − sin x}{1 − sin^{2} x} dx = ∫ \frac{dx}{cos^{2} x} − ∫ \frac{sin x}{cos^{2} x} dx = cos x = t − sin x dx = dt = tg x + ∫ \frac{dt}{t^{2}} =

= tg x + \frac{t^{−1}}{−1} + c = tg x\frac{1}{t} + c = tg x − \frac{1}{cos x} + c = \frac{sin x}{cos x} − \frac{1}{cos x} + c = \frac{sin x − 1}{cos x} + c, for x∈R − \{\frac{π}{2} + kπ; k ∈Z\}.

∫ \frac{sin x dx}{\sqrt{cos^{5} x}} = cos x = t − sin x dx = dt = ∫ \frac{− dt}{\sqrt{t^{5}}} = − ∫ t^{− \frac{5}{2}} dt = −\frac{t^{− \frac{3}{2}}}{−\frac{3}{2}} + c = \frac{2}{3} \frac{1}{\sqrt{t^{3}}} + c = \frac{2}{3} \frac{1}{\sqrt{cos^{3} x}} + c, for x∈ ⋃_{k∈Z} (−\frac{π}{2} + 2kπ; \frac{π}{2} + 2kπ).

∫ \frac{cos x dx}{\sqrt[3]{sin^{2} x}} = sin x = t ; cos x dx = dt = ∫ \frac{dt}{\sqrt[3]{t^{2}}} = ∫ t^{− \frac{2}{3}} dt = \frac{t^{\frac{1}{3}}}{\frac{1}{3}} + c = 3\sqrt[3]{t} + c = 3\sqrt[3]{sin x} + c,  for x∈ ⋃_{k∈Z} (0 + 2kπ; π + 2kπ).

∫ sin^{2} x dx = ∫ \frac{1 − cos 2x}{2} dx = \frac{x}{2} − \frac{sin 2x}{4} + c, for x∈R.

∫ cos^{2} x dx = ∫ \frac{1 + cos 2x}{2} dx = \frac{x}{2} + \frac{sin 2x}{4} + c, for x∈R.

∫ sin^{3} x dx = ∫ (1 − cos^{2} x) sin x dx = cos x = t − sin x dx = dt = − ∫ (1 − t^{2}) dt = −t + \frac{t^{3}}{3} + c = \frac{cos^{3} x}{3} − cos x + c,  for x∈R.

∫ cos^{3} x dx = ∫ (1 − sin^{2} x) sin x dx = sin x = t ; cos x dx = dt = ∫ (1 − t^{2}) dt = t − \frac{t^{3}}{3} + c = sin x − \frac{sin^{3} x}{3} + c,  for x∈R.

∫ sin^{n} x dx = ∫ sin x sin^{n−1} x dx = u = sin^{n−1} x ⇒ u′ = (n − 1) sin^{n−2} x cos x v′ = sin x ⇒ v = − cos x = − cos x sin^{n−1} x + (n − 1) ∫ sin^{n−2} x cos^{2} x dx =

= − cos x sin^{n−1} x + (n − 1) ∫ sin^{n−2} x(1 − sin^{2} x) dx = − cos x sin^{n−1} x + (n − 1) ∫ sin^{n−2} x dx − (n − 1) ∫ sin^{n} x dx

⟹^{(i.e.equation)} − cos x sin^{n−1} x + (n − 1) ∫ sin^{n−2} x dx = ∫ sin^{n} x dx + (n − 1) ∫ sin^{n} x dx = n ∫ sin^{n} x dx

⟹ ∫ sin^{n} x dx = −\frac{cos x sin^{n−1} x}{n} + \frac{n − 1}{n} ∫ sin^{n−2} x dx, for n = 3, 4, 5, …, x∈R.

∫ sin^{2k+1} x dx = ∫ sin x sin^{2k} x dx = ∫ sin x(sin^{2} x)^{k} dx = ∫ sin x(1 − cos^{2} x)^{k} dx = cos x = t − sin x dx = dt = − ∫ (1 − t^{2})^{k} dt =

= −∫ [∑_{j=0}^{k} \binom{k}{j}(−t^{2})^{j}]dt = −∫ [∑_{j=0}^{k} \binom{k}{j}(−1)^{j}t^{2j}]dt = ∑_{j=0}^{k} \binom{k}{j}(−1)^{j+1} \frac{t^{2j+1}}{2j + 1} + c = ∑_{j=0}^{k} \binom{k}{j}(−1)^{j+1} \frac{cos^{2j+1} x}{2j + 1} + c =

= − cos x + \binom{k}{1}\frac{cos^{3} x}{3} − ⋯ + \binom{k}{k − 1}(−1)^{k} \frac{cos^{2k−1} x}{2k − 1} + (−1)^{k+1} \frac{cos^{2k+1} x}{2k + 1} + c, for k ∈N, x∈R.

∫ tg^{2} x dx = ∫ \frac{sin^{2} x}{cos^{2} x} dx = ∫ \frac{1 − cos^{2} x}{cos^{2} x} dx = ∫ [\frac{1}{cos^{2} x} − 1] dx = tg x − x + c, for x∈R − \{\frac{π}{2} + kπ; k ∈Z\}.

∫ cos^{n} x dx = ∫ cos x cos^{n−1} x dx = u = cos^{n−1} x ⇒ u′ = −(n − 1) cos^{n−2} x sin x v′ = cos x ⇒ v = sin x = sin x cos^{n−1} x + (n − 1) ∫ cos^{n−2} x sin^{2} x dx =

= sin x cos^{n−1} x + (n − 1) ∫ cos^{n−2} x(1 − cos^{2} x) dx = sin x cos^{n−1} x + (n − 1) ∫ cos^{n−2} x dx − (n − 1) ∫ cos^{n} x dx

⟹^{(i.e.equation)} sin x cos^{n−1} x + (n − 1) ∫ cos^{n−2} x dx = ∫ cos^{n} x dx + (n − 1) ∫ cos^{n} x dx = n ∫ cos^{n} x dx

⟹ ∫ cos^{n} x dx = \frac{sin x cos^{n−1} x}{n} + \frac{n − 1}{n} ∫ cos^{n−2} x dx, for n = 3, 4, 5, …, x∈R.

∫ cos^{2k+1} x dx = ∫ cos x cos^{2k} x dx = ∫ cos x(cos^{2} x)^{k} dx = ∫ cos x(1 − sin^{2} x)^{k} dx = sin x = t ; cos x dx = dt = ∫ (1 − t^{2})^{k} dt =

= ∫ [∑_{j=0}^{k} \binom{k}{j}(−t^{2})^{j}] dt = ∫ [∑_{j=0}^{k} \binom{k}{j}(−1)^{j}t^{2j}] dt = ∑_{j=0}^{k} \binom{k}{j}(−1)^{j} \frac{t^{2j+1}}{2j + 1} + c = ∑_{j=0}^{k} \binom{k}{j}(−1)^{j} \frac{sin^{2j+1} x}{2j + 1} + c =

= sin x − \binom{k}{1}\frac{sin^{3} x}{3} − ⋯ + \binom{k}{k − 1}(−1)^{k−1} \frac{sin^{2k−1} x}{2k − 1} + (−1)^{k} \frac{sin^{2k+1} x}{2k + 1} + c, for k ∈N, x∈R.

∫ \frac{sin x − cos x}{\sqrt[4]{sin x + cos x}} dx = t = tg \frac{x}{2}, dx = \frac{2 dt}{1 + t^{2}} cos x = \frac{1 − t^{2}}{1 + t^{2}} , sin x = \frac{2t}{1 + t^{2}} = ∫ \frac{\frac{2t}{1 + t^{2}} − \frac{1 − t^{2}}{1 + t^{2}}}{\sqrt[4]{\frac{2t}{1 + t^{2}} + \frac{1 − t^{2}}{1 + t^{2}}}} \frac{2 dt}{1 + t^{2}} = ∫ (t^{2} + 2t − 1)\sqrt[4]{\frac{1 + t^{2}}{1 + 2t − t^{2}}} dt = · · ·.  ^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}^{.}^{.}^{.}^{.}^{.}_{.}^{.}^{.}^{.}^{.}^{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.} _{.}^{.}^{.}^{.}^{.}^{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}^{.}^{.}^{.}^{.}^{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.} _{.}^{.}^{.}^{.}^{.}^{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.} _{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}^{^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}} ^{^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}}^{^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}} ^{^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}}^{^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}} ^{^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}}^{^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}} ^{^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}}_{. . . . . . . . . . . . . . . .}_{. . . . . . . . . . . . . . . .}

∫ \frac{sin x − cos x}{\sqrt[4]{sin x + cos x}} dx = sin x + cos x = t (cos x − sin x) dx = dt = ∫ \frac{− dt}{\sqrt[4]{t}} = − ∫ t^{− \frac{1}{4}} dt = −\frac{t^{\frac{3}{4}}}{\frac{3}{4}} + c = c − \frac{4}{3}\sqrt[4]{(sin x + cos x)^{3}},

for x∈ ⋃_{k∈Z} (\frac{π}{2} + 2kπ; \frac{3π}{2} + 2kπ).

∫ \frac{x^{2}}{sin x^{3}} dx = \begin{matrix}t = x^{3} \\ dt = 3x^{2} dx \end{matrix} = \frac{1}{3} ∫ \frac{dt}{sin t} = u = tg \frac{t}{2}, dt = \frac{2 du}{1 + u^{2}} , sin t = \frac{2u}{1 + u^{2}} = \frac{1}{3} ∫ \frac{1 + u^{2}}{2u} \frac{2 du}{1 + u^{2}} = \frac{1}{3} ∫ \frac{du}{u} =

= \frac{1}{3} ln |u| + c = \frac{1}{3} ln |tg \frac{t}{2}| + c = ln \sqrt[3]{|tg \frac{x^{3}}{2} |} + c, for x∈R − \{\sqrt[3]{kπ}; k ∈Z\}.

∫ \frac{dx}{sin^{2} x cos^{2} x} = ∫ \frac{4 dx}{(2 sin x cos x)^{2}} = ∫ \frac{4 dx}{sin^{2} 2x} = \begin{matrix}2x = d \\ 2 dx = dt \end{matrix} = ∫ \frac{2 dt}{sin^{2} t} = −2 cotg t + c = −2 cotg 2x + c, for x∈R − \{\frac{kπ}{2} ; k ∈Z\}.

∫ \frac{dx}{sin^{2} x cos^{2} x} = ∫ \frac{sin^{2} x + cos^{2} x}{sin^{2} x cos^{2} x} dx = ∫ \frac{dx}{sin^{2} x} + ∫ \frac{dx}{cos^{2} x} = tg x − cotg x + c =

= \frac{sin x}{cos x} − \frac{cos x}{sin x} + c = 2\frac{sin^{2} x − cos^{2} x}{2 sin x cos x} + c = 2\frac{− cos 2x}{sin 2x} + c = −2 cotg 2x + c, for x∈R − \{\frac{kπ}{2} ; k ∈Z\}.

∫ cos ax sin bx dx = u = sin bx ⇒ u′ = b cos bx v′ = cos ax ⇒ v = \frac{sin ax}{a} = \frac{sin ax sin bx}{a} − \frac{b}{a} ∫ sin ax cos bx dx = u = cos bx ⇒ u′ = −b sin bx v′ = sin ax ⇒ v = −\frac{cos ax}{a} =

= \frac{sin ax sin bx}{a} − \frac{b}{a} [−\frac{cos ax cos bx}{a} − \frac{b}{a} ∫ cos ax sin bx dx] = \frac{sin ax sin bx}{a} + \frac{b}{a^{2}} cos ax cos bx + \frac{b^{2}}{a^{2}} ∫ cos ax sin bx dx

⟹^{(i.e.equation)} \frac{sin ax sin bx}{a} + \frac{b}{a^{2}} cos ax cos bx + c = ∫ cos ax sin bx dx − \frac{b^{2}}{a^{2}}∫ cos ax sin bx dx = \frac{a^{2}−b^{2}}{a^{2}}∫ cos ax sin bx dx

⟹ ∫ cos ax sin bx dx = \frac{a}{a^{2} − b^{2}} sin ax sin bx + \frac{b}{a^{2} − b^{2}} cos ax cos bx + c, for a, b∈R − \{0\}, a≠b, x∈R.

∫ cos ax sin bx dx = u = cos ax ⇒ u′ = −a sin ax v′ = sin bx ⇒ v = −\frac{cos bx}{b} = −\frac{cos ax cos bx}{b} − \frac{a}{b} ∫ sin ax cos bx dx = u = sin ax ⇒ u′ = a cos ax v′ = cos bx ⇒ v = \frac{sin bx}{b} =

= −\frac{cos ax cos bx}{b} − \frac{a}{b} [\frac{sin ax sin bx}{b} − \frac{a}{b} ∫ cos ax sin bx dx] = −\frac{cos ax cos bx}{b} − \frac{a}{b^{2}} sin ax sin bx + \frac{a^{2}}{b^{2}} ∫ cos ax sin bx dx

⟹^{(i.e.equation)} c− \frac{cos ax cos bx}{b} − \frac{a}{b^{2}} sin ax sin bx = ∫ cos ax sin bx dx− \frac{a^{2}}{b^{2}}∫ cos ax sin bx dx = \frac{b^{2}−a^{2}}{b^{2}} ∫ cos ax sin bx dx

⟹ ∫ cos ax sin bx dx = \frac{b}{a^{2} − b^{2}} cos ax cos bx + \frac{a}{a^{2} − b^{2}} sin ax sin bx + c, for a, b∈R − \{0\}, a≠b, x∈R.

∫ cos ax sin ax dx = ∫ \frac{sin 2ax}{2} dx = \frac{− cos 2ax}{2 · 2a} + c = c − \frac{cos 2ax}{4a}, for a = b, a∈R − \{0\}, x∈R.

∫ x ln^{2} x dx = \begin{matrix}u = ln^{2} x ⇒ & u′ = \frac{2 ln x}{x} \\ v′ = x ⇒ & v = \frac{x^{2}}{2} \end{matrix} = \frac{x^{2}}{2} ln^{2} x − ∫ \frac{x^{2}}{2} \frac{2 ln x}{x} dx = \frac{x^{2}}{2} ln^{2} x − ∫ x ln x dx = \begin{matrix}u = ln x ⇒ & u′ = \frac{1}{x} \\ v′ = x ⇒ & v = \frac{x^{2}}{2} \end{matrix} =

= \frac{x^{2}}{2} ln^{2} x − [\frac{x^{2}}{2} ln x − ∫ \frac{x^{2}}{2} \frac{1}{x} dx] = \frac{x^{2}}{2} ln^{2} x − \frac{x^{2}}{2} ln x + ∫ \frac{x dx}{2} = \frac{x^{2}}{2} ln^{2} x − \frac{x^{2}}{2} ln x + \frac{x^{2}}{4} + c, for x > 0.

∫ \frac{dx}{(1 − x)x^{2}} = ∫ [\frac{1}{x − 1} − \frac{1}{x^{2}} − \frac{1}{x}] dx = ln |x − 1| − \frac{x^{−1}}{−1} − ln |x| + c = ln |\frac{x − 1}{x}| + \frac{1}{x} + c, for x∈R − \{0, 1\}.

\frac{1}{(1 − x)x^{2}} = \frac{A}{x − 1} + \frac{B}{x^{2}} + \frac{C}{x} ⟹ 1 = Ax^{2} + B(x − 1) + Cx(x − 1) ⟹ A = 1, B = −1, C = −1

∫ ln x dx = \begin{matrix}u = ln x ⇒ & u′ = \frac{1}{x} \\ v′ = 1 ⇒ & v = x \end{matrix} = x ln x − ∫ x\frac{dx}{x} = x ln x − ∫ dx = x ln x − x + c, for x > 0.

∫ \sqrt{a^{2} − x^{2}} dx = x = a sin t dx = a cos t dt = ∫ a^{2} cos^{2} t dt = a^{2} ∫ \frac{1 + cos 2t}{2} dt = \frac{a^{2}}{2} t + a^{2} \frac{sin 2t}{4} + c_{1} = \frac{a^{2}}{2} t + a^{2} \frac{2 sin t cos t}{4} + c_{1} =

t = arcsin \frac{x}{a}, \sqrt{a^{2} − x^{2}} = \sqrt{a^{2} − a^{2} sin^{2} t} = |a| \sqrt{1 − sin^{2} t} = a\sqrt{cos^{2} t} = a |cos t| = a cos t, x∈(−a; a) ⇒ t∈(−\frac{π}{2} ; \frac{π}{2} )

= \frac{a^{2}}{2} t + \frac{a sin t · a cos t}{2} + c_{1} = \frac{a^{2}}{2} arcsin \frac{x}{a} + \frac{x\sqrt{a^{2} − x^{2}}}{2} + c_{1}, for a > 0, x∈(−a; a).

∫ \sqrt{a^{2} − x^{2}} dx = ∫ \sqrt{(−a)^{2} − x^{2}} dx = \frac{a^{2}}{2} arcsin \frac{x}{−a} + \frac{x\sqrt{a^{2} − x^{2}}}{2} + c_{1}, for a < 0 [i.e. −a > 0, (−a)^{2} = a^{2}], x∈(a; −a).

⟹ ∫ \sqrt{a^{2} − x^{2}} dx = \frac{a^{2}}{2} arcsin \frac{x}{|a|} + \frac{x\sqrt{a^{2} − x^{2}}}{2} + c_{1}, for a∈R − \{0\}, x∈(− |a|; |a|).

∫ \sqrt{a^{2} − x^{2}} dx = ∫ \frac{a^{2} − x^{2}}{\sqrt{a^{2} − x^{2}}} dx = ∫ \frac{a^{2} dx}{\sqrt{a^{2} − x^{2}}} − ∫ \frac{x^{2} dx}{\sqrt{a^{2} − x^{2}}} = \begin{matrix}u = x ⇒ & u′ = 1 \\ v′ = \frac{x}{\sqrt{a^{2} − x^{2}}} ⇒ & v = −\sqrt{a^{2} − x^{2}} \end{matrix} =

= ∫ \frac{a^{2} dx}{\sqrt{a^{2} − x^{2}}} − [−x\sqrt{a^{2} − x^{2}} + ∫ \sqrt{a^{2} − x^{2}} dx] = a^{2} arcsin \frac{x}{|a|} + x\sqrt{a^{2} − x^{2}} − ∫ \sqrt{a^{2} − x^{2}} dx

⟹^{(i.e.equation)} a^{2} arcsin \frac{x}{|a|} + x\sqrt{a^{2} − x^{2}} + c_{0} = ∫ \sqrt{a^{2} − x^{2}} dx + ∫ \sqrt{a^{2} − x^{2}} dx = 2 ∫ \sqrt{a^{2} − x^{2}} dx

⟹ ∫ \sqrt{a^{2} − x^{2}} dx = \frac{a^{2}}{2} arcsin \frac{x}{|a|} + \frac{x\sqrt{a^{2} − x^{2}}}{2} + c_{1}, for a∈R − \{0\}, x∈(− |a|; |a|).

∫ \sqrt{a^{2} − x^{2}} dx = ∫ \frac{a(1 − t^{2})}{1 + t^{2}} \frac{2a(1 − t^{2})}{(1 + t^{2})^{2}} dt = 2a^{2} ∫ \frac{(1 − t^{2})^{2}}{(1 + t^{2})^{3}} dt = 2a^{2} ∫ \frac{1 − 2t^{2} + t^{4}}{(1 + t^{2})^{3}} dt = 2a^{2} ∫ [\frac{1 + 2t^{2} + t^{4} − 4t^{2}}{(1 + t^{2})^{3}} ] dt =

x = \frac{2at}{1 + t^{2}} , dx = \frac{2a(1 + t^{2}) − 2at · 2t}{(1 + t^{2})^{2}} dt = \frac{2a(1 − t^{2})}{(1 + t^{2})^{2}} dt, a − xt = a − \frac{2at^{2}}{1 + t^{2}} = \frac{a + at^{2} − 2at^{2}}{1 + t^{2}} = \frac{a(1 − t^{2})}{1 + t^{2}}

= 2a^{2} ∫ [\frac{(1 + t^{2})^{2} − 4t^{2} − 4 + 4}{(1 + t^{2})^{3}}] dt = 2a^{2} ∫ [\frac{(1 + t^{2})^{2} − 4(1 + t^{2}) + 4}{(1 + t^{2})^{3}}] dt = 2a^{2} ∫ [\frac{1}{1 + t^{2}} − \frac{4}{(1 + t^{2})^{2}} + \frac{4}{(1 + t^{2})^{3}} ] dt =

= 2a^{2} [arctg t − 4 (\frac{1}{2} arctg t + \frac{1}{2} \frac{t}{1 + t^{2}} ) + 4 (\frac{3}{8} arctg t + \frac{3}{8} \frac{t}{1 + t^{2}} + \frac{1}{4} \frac{t}{(1 + t^{2})^{2}} )]+c_{2} = a^{2} arctg t− \frac{a^{2}t}{1 + t^{2}} + \frac{2a^{2}t}{(1 + t^{2})^{2}} +c_{2} =

= x = \frac{2at}{1 + t^{2}} , \sqrt{a^{2} − x^{2}} = a − xt ⇒ t = \frac{a − \sqrt{a^{2} − x^{2}}}{x} = a^{2} arctg t − \frac{a · 2at}{2(1 + t^{2})} + \frac{4a^{2}t^{2}}{2t(1 + t^{2})^{2}} + c_{2} =

= a^{2} arctg \frac{a − \sqrt{a^{2} − x^{2}}}{x} − \frac{ax}{2} + \frac{x^{3}}{2 (a − \sqrt{a^{2} − x^{2}})} + c_{2} = a^{2} arctg \frac{a − \sqrt{a^{2} − x^{2}}}{x} − \frac{ax}{2} + \frac{x^{3}}{2 (a − \sqrt{a^{2} − x^{2}})} \frac{a + \sqrt{a^{2} − x^{2}}}{a + \sqrt{a^{2} − x^{2}}} + c_{2} =

= a^{2} arctg \frac{a − \sqrt{a^{2} − x^{2}}}{x} − \frac{ax}{2} + \frac{x}{2} (a + \sqrt{a^{2} − x^{2}}) + c_{2} = a^{2} arctg \frac{a − \sqrt{a^{2} − x^{2}}}{x} + \frac{x\sqrt{a^{2} − x^{2}}}{2} + c_{2},

for a∈R − {0}, x∈(− |a|; |a|).

2-nd Euler: \sqrt{a^{2} − x^{2}} = a − xt ⇒ a^{2} − x^{2} = x^{2}t^{2} − 2axt + a^{2} ⇒ 2axt = x^{2} + x^{2}t^{2} ⇒ x = 0 or 2at = x + xt^{2}

∫ \sqrt{a^{2} + x^{2}} dx = x = a sinh t dx = a cosh t dt = ∫ a^{2} cosh^{2} t dt = a^{2} ∫ \frac{(e^{t} + e^{−t})^{2}}{4} dt = a^{2} ∫ \frac{e^{2t} +2 + e^{−2t}}{4} dt = a^{2} [\frac{e^{2t}}{8} + \frac{t}{2} − \frac{e^{−2t}}{8} ] + c =

\sqrt{a^{2} + x^{2}} = \sqrt{a^{2} + a^{2} sinh^{2} t} = |a| \sqrt{1 + sinh^{2} t} = a\sqrt{cosh^{2} t} = a |cosh t| = a cosh t, t∈R

Let u =: e^{t} > 0 ⟹ x = a sinh t = \frac{a}{2}(e^{t} − e^{−1}) = \frac{a}{2}(u − u^{−1}) ⟹ \frac{2x}{a} = u − u^{−1} ⟹ u^{2} − \frac{2x}{a}u − 1 = 0 ⟹

u_{1,2} = \frac{\frac{2x}{a} ± \sqrt{\frac{4x^{2}}{a^{2}} + 4}}{2} = \frac{\frac{2x}{a} ± \sqrt{\frac{4x^{2}+4a^{2}}{a^{2}}}}{2} = \frac{\frac{2x}{a} ± \frac{2}{a} \sqrt{a^{2} + x^{2}}}{2} = \frac{x ± \sqrt{a^{2} + x^{2}}}{a} ⟹ e^{t} = \frac{x + \sqrt{a^{2} + x^{2}}}{a} ⟹

e^{−2t} = \frac{a^{2}}{(x + \sqrt{a^{2} + x^{2}})^{2}} = \frac{a^{2}}{(x + \sqrt{a^{2} + x^{2}})^{2}} \begin{matrix}(x − \sqrt{a^{2} + x^{2}}) & 2 \\ (x − \sqrt{a^{2} + x^{2}}) & 2 \end{matrix} = \frac{a^{2} (x − \sqrt{a^{2} + x^{2}})^{2}}{[x^{2} − (a^{2} + x^{2})]^{2}} = \frac{(x − \sqrt{a^{2} + x^{2}})^{2}}{a^{2}}

\sqrt{a^{2} + x^{2}} > \sqrt{x^{2}} ≥ x ⟹ \sqrt{a^{2} + x^{2}} − x > 0 ⟹ t = ln \frac{x + \sqrt{a^{2} + x^{2}}}{a} = ln (x + \sqrt{a^{2} + x^{2}}) − ln a

= \frac{(x + \sqrt{a^{2} + x^{2}})^{2}}{8} + a^{2} \frac{ln (x + \sqrt{a^{2} + x^{2}}) − ln a}{2} − \frac{(x − \sqrt{a^{2} + x^{2}})^{2}}{8} + c_{1} =

= \frac{x^{2} + 2x\sqrt{a^{2} + x^{2}} + a^{2} + x^{2}}{8} − \frac{x^{2} − 2x\sqrt{a^{2} + x^{2}} + a^{2} + x^{2}}{8} + a^{2} \frac{ln (x + \sqrt{a^{2} + x^{2}})}{2} − \frac{a^{2}}{2} ln a + c_{1} =

= \frac{4x\sqrt{a^{2} + x^{2}}}{8} + \frac{a^{2}}{2} ln (x + \sqrt{a^{2} + x^{2}}) + c = \frac{x\sqrt{a^{2} + x^{2}}}{2} + \frac{a^{2}}{2} ln (x + \sqrt{a^{2} + x^{2}}) + c, for a > 0, x∈R.

∫ \sqrt{a^{2} + x^{2}} dx = ∫ \sqrt{(−a)^{2} + x^{2}} dx = \frac{x\sqrt{a^{2} + x^{2}}}{2} + \frac{a^{2}}{2} ln (x + \sqrt{a^{2} + x^{2}}) + c, for a < 0 [i.e. −a > 0, (−a)^{2} = a^{2}], x∈R.

⟹ ∫ \sqrt{a^{2} + x^{2}} dx = \frac{x\sqrt{a^{2} + x^{2}}}{2} + \frac{a^{2}}{2} ln (x + \sqrt{a^{2} + x^{2}}) + c, for a∈R − \{0\}, x∈R.

∫ \sqrt{a^{2} + x^{2}} dx = ∫ \frac{t^{2} + a^{2}}{2t} \frac{t^{2} + a^{2}}{2t^{2}} dt = ∫ \frac{t^{4} + 2t^{2}a^{2} + a^{4}}{4t^{3}} dt = \frac{1}{4} ∫ [t + 2a^{2}t^{−1} + a^{4}t^{−3}] dt = \frac{1}{4} [\frac{t^{2}}{2} + 2a^{2} ln |t| + a^{4} \frac{t^{−2}}{−2}] + c =

1-st Euler: \sqrt{a^{2} + x^{2}} = t − x ⟹ a^{2} + x^{2} = t^{2} − 2tx + x^{2} ⟹ x = \frac{t^{2} − a^{2}}{2t} , t = x + \sqrt{a^{2} + x^{2}}

\sqrt{a^{2} + x^{2}} = t − \frac{t^{2} − a^{2}}{2t} = \frac{t^{2} + a^{2}}{2t} , dx = \frac{2t · 2t − 2(t^{2} − a^{2})}{4t^{2}} dt = \frac{2t^{2} + 2a^{2}}{4t^{2}} dt = \frac{t^{2} + a^{2}}{2t^{2}} dt

t^{2} = (x + \sqrt{a^{2} + x^{2}})^{2} , t^{−2} = \frac{1}{(x + \sqrt{a^{2} + x^{2}})^{2}} \begin{matrix}(x − \sqrt{a^{2} + x^{2}}) & 2 \\ (x − \sqrt{a^{2} + x^{2}}) & 2 \end{matrix} = \begin{matrix}(x − \sqrt{a^{2} + x^{2}}) & 2 \\ [x^{2} − (a^{2} + x^{2})] & 2 \end{matrix} = \frac{(x − \sqrt{a^{2} + x^{2}})^{2}}{a^{4}}

= \frac{(x + \sqrt{a^{2} + x^{2}})^{2}}{8} − \frac{(x − \sqrt{a^{2} + x^{2}})^{2}}{8} + \frac{a^{2}}{2} ln |x + \sqrt{a^{2} + x^{2}}| + c =

= \frac{x^{2} + 2x\sqrt{a^{2} + x^{2}} + a^{2} + x^{2}}{8} − \frac{x^{2} − 2x\sqrt{a^{2} + x^{2}} + a^{2} + x^{2}}{8} + \frac{a^{2}}{2} ln (x + \sqrt{a^{2} + x^{2}}) + c =

= \frac{4x\sqrt{a^{2} + x^{2}}}{8} + \frac{a^{2}}{2} ln (x + \sqrt{a^{2} + x^{2}}) + c = \frac{x\sqrt{a^{2} + x^{2}}}{2} + \frac{a^{2}}{2} ln (x + \sqrt{a^{2} + x^{2}}) + c, for a∈R − \{0\}, x∈R.

∫ \sqrt{a^{2} + x^{2}} dx = ∫ \frac{a^{2} + x^{2}}{\sqrt{a^{2} + x^{2}}} dx = ∫ \frac{a^{2} dx}{\sqrt{a^{2} + x^{2}}} + ∫ \frac{x^{2} dx}{\sqrt{a^{2} + x^{2}}} = \begin{matrix}u = x ⇒ & u′ = 1 \\ v′ = \frac{x}{\sqrt{a^{2} + x^{2}}} ⇒ & v = \sqrt{a^{2} + x^{2}} \end{matrix} =

= ∫ \frac{a^{2} dx}{\sqrt{a^{2} + x^{2}}} + [x\sqrt{a^{2} + x^{2}} − ∫ \sqrt{a^{2} + x^{2}} dx] = a^{2} ln (x + \sqrt{a^{2} + x^{2}}) + x\sqrt{a^{2} + x^{2}} − ∫ \sqrt{a^{2} + x^{2}} dx

⟹^{(i.e.equation)} a^{2} ln (x + \sqrt{a^{2} + x^{2}}) + x\sqrt{a^{2} + x^{2}} + c_{0} = ∫ \sqrt{a^{2} + x^{2}} dx + ∫ \sqrt{a^{2} + x^{2}} dx = 2 ∫ \sqrt{a^{2} + x^{2}} dx

⟹ ∫ \sqrt{a^{2} + x^{2}} dx = \frac{a^{2}}{2} ln (x + \sqrt{a^{2} + x^{2}}) + \frac{x\sqrt{a^{2} + x^{2}}}{2} + c, for a∈R − \{0\}, x∈R.

∫ \sqrt{a^{2} + x^{2}} dx = ∫ \frac{a^{2} dt}{cos^{3} t} = ∫ \frac{a^{2} cos t dt}{cos^{4} t} = ∫ \frac{a^{2} cos t dt}{(1 − sin^{2} t)^{2}} = u = sin t ; du = cos t dt = ∫ \frac{a^{2} du}{(1 − u^{2})^{2}} = ∫ \frac{a^{2} du}{(u^{2} − 1)^{2}} =

x = a tg t ⟹ dx = \frac{a dt}{cos^{2} t} ⟹ \sqrt{a^{2} + x^{2}} = \sqrt{a^{2} + \frac{a^{2} sin^{2} t}{cos^{2} t}} = \sqrt{\frac{a^{2} cos^{2} t + a^{2} sin^{2} t}{cos^{2} t}} = \sqrt{\frac{a^{2}}{cos^{2} t}} = \frac{a}{cos t}

x∈R ⟹ t = arctg x∈(−\frac{π}{2} ; \frac{π}{2} ) ⟹ \sqrt{cos^{2} x} = |cos x| = cos x, tg^{2} t = \frac{sin^{2} t}{cos^{2} t} = \frac{sin^{2} t}{1 − sin^{2} t} ⟹ sin^{2} t = \frac{tg^{2} t}{1 + tg^{2} t}

sgn sin t = sgn tg t ⟹ sin t = \frac{tg t}{\sqrt{1 + tg^{2} t}} = \frac{\frac{x}{a}}{\sqrt{1 + \frac{x^{2}}{a^{2}}}} = \frac{x}{\sqrt{a^{2} + x^{2}}} ⟹ sin t ± 1 = \frac{x ± \sqrt{a^{2} + x^{2}}}{\sqrt{a^{2} + x^{2}}}

= ∫ \frac{a^{2} du}{(u − 1)^{2}(u + 1)^{2}} = \frac{a^{2}}{4} ∫ [\frac{1}{u + 1} − \frac{1}{u − 1} + \frac{1}{(u + 1)^{2}} + \frac{1}{(u − 1)^{2}} ] du = \begin{matrix}u − 1 = r \\ du = dr \end{matrix} \begin{matrix}u + 1 = s \\ du = ds \end{matrix} =

\frac{1}{(u − 1)^{2}(u + 1)^{2}} = \frac{A}{u + 1} + \frac{B}{(u + 1)^{2}} + \frac{C}{u − 1} + \frac{D}{(u − 1)^{2}} ⟹ A = \frac{1}{4}, B = \frac{1}{4}, C = −\frac{1}{4}, D = \frac{1}{4}

= \frac{a^{2}}{4} [ln |u + 1| − ln |u − 1| + ∫ r^{−2} dr + ∫ s^{−2} ds] = \frac{a^{2}}{4} ln |\frac{u + 1}{u − 1}| + \frac{a^{2}}{4} [\frac{r^{−1}}{−1} + \frac{s^{−1}}{−1} ] + c_{1} =

= \frac{a^{2}}{4} ln |\frac{u + 1}{u − 1}| − \frac{a^{2}}{4} [\frac{1}{u − 1} + \frac{1}{u + 1}] + c_{1} = \frac{a^{2}}{4} ln |\frac{u + 1}{u − 1}| − \frac{2ua^{2}}{4(u^{2} − 1)} + c_{1} = \frac{a^{2}}{4} ln |\frac{u + 1}{u − 1}| − \frac{ua^{2}}{2(u^{2} − 1)} + c_{1} =

= \frac{a^{2}}{4} ln |\frac{sin t + 1}{sin t − 1}| − \frac{a^{2} sin t}{2(sin^{2} t − 1)} + c_{1} = \frac{a^{2}}{4} ln |\frac{x + \sqrt{a^{2} + x^{2}}}{x − \sqrt{a^{2} + x^{2}}} | − \frac{xa^{2}}{2\sqrt{a^{2} + x^{2}}} [\frac{x^{2}}{a^{2} + x^{2}} − 1]^{−1} + c_{1} =

= \frac{a^{2}}{4} ln |\frac{x + \sqrt{a^{2} + x^{2}}}{x − \sqrt{a^{2} + x^{2}}} · \frac{x + \sqrt{a^{2} + x^{2}}}{x + \sqrt{a^{2} + x^{2}}} | − \frac{xa^{2}}{2\sqrt{a^{2} + x^{2}}} \frac{a^{2} + x^{2}}{−a^{2}} + c_{1} = \frac{a^{2}}{4} ln |\frac{(x + \sqrt{a^{2} + x^{2}})^{2}}{x^{2} − (a^{2} + x^{2})} | + \frac{x\sqrt{a^{2} + x^{2}}}{2} + c_{1} =

= \frac{a^{2}}{4} ln |\frac{(x + \sqrt{a^{2} + x^{2}})^{2}}{a^{2}} | + \frac{x\sqrt{a^{2} + x^{2}}}{2} + c_{1} = \frac{a^{2}}{2} ln |x + \sqrt{a^{2} + x^{2}}| − \frac{a^{2}}{2} ln a + \frac{x\sqrt{a^{2} + x^{2}}}{2} + c_{1} =

= \frac{a^{2}}{2} ln (x + \sqrt{a^{2} + x^{2}}) + \frac{x\sqrt{a^{2} + x^{2}}}{2} + c, for a > 0, x∈R.

∫ \sqrt{a^{2} + x^{2}} dx = ∫ \sqrt{(−a)^{2} + x^{2}} dx = \frac{x\sqrt{a^{2} + x^{2}}}{2} + \frac{a^{2}}{2} ln (x + \sqrt{a^{2} + x^{2}}) + c, for a < 0 [i.e. −a > 0, (−a)^{2} = a^{2}], x∈R.

⟹ ∫ \sqrt{a^{2} + x^{2}} dx = \frac{x\sqrt{a^{2} + x^{2}}}{2} + \frac{a^{2}}{2} ln (x + \sqrt{a^{2} + x^{2}}) + c, for a∈R − \{0\}, x∈R.

∫ \sqrt{x^{2} − a^{2}} dx = x = a cosh t dx = a sinh t dt = ∫ a^{2} sinh^{2} t dt = a^{2} ∫ \frac{(e^{t} − e^{−t})^{2}}{4} dt = a^{2} ∫ \frac{e^{2t} −2 + e^{−2t}}{4} dt = a^{2} [\frac{e^{2t}}{8} − \frac{t}{2} − \frac{e^{−2t}}{8} ] + c =

\sqrt{x^{2} − a^{2}} = \sqrt{a^{2} cosh^{2} t − a^{2}} = a\sqrt{cosh^{2} t − 1} = a\sqrt{sinh^{2} t} = a |sinh t| = a sinh t, t > 0, x > a

u =: e^{t} > 0 ⟹ x = a cosh t = \frac{a}{2}(e^{t} + e^{−1}) = \frac{a}{2}(u + u^{−1}) ⟹ \frac{2x}{a} = u + u^{−1} ⟹ u^{2} − \frac{2x}{a}u + 1 = 0 ⟹

u_{1,2} = \frac{\frac{2x}{a} ± \sqrt{\frac{4x^{2}}{a^{2}} − 4}}{2} = \frac{\frac{2x}{a} ± \sqrt{\frac{4x^{2}−4a^{2}}{a^{2}}}}{2} = \frac{\frac{2x}{a} ± \frac{2}{a} \sqrt{x^{2} − a^{2}}}{2} = \frac{x ± \sqrt{x^{2} − a^{2}}}{a} > 0 ⟹

e^{t} = \frac{x − \sqrt{x^{2} − a^{2}}}{a} ⇒ |t| = ln |\frac{x − \sqrt{x^{2} − a^{2}}}{a} | = ln |\frac{x^{2} − (x^{2} − a^{2})}{a (x + \sqrt{x^{2} − a^{2}})}| = ln |\frac{a}{x + \sqrt{x^{2} − a^{2}}} | < ln |\frac{a}{a}| = ln 1 = 0 ^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}^{.}^{.}^{.}^{.}_{.}^{.}^{.}^{.}^{.}^{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}^{.}^{.}^{.}^{.}^{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}^{.}^{.}^{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}^{.}^{.}^{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}^{^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}}^{^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}}^{^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}} ^{^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}}^{^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}} ^{^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}}^{^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}} ^{^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}}_{. . . . . . . . . .}_{. . . . . . . . . .}

e^{t} = \frac{x + \sqrt{x^{2} − a^{2}}}{a} ⇒ t = ln (\frac{x + \sqrt{x^{2} − a^{2}}}{a} ) ⇒ e^{−2t} = \frac{a^{2}}{(x + \sqrt{x^{2} − a^{2}})^{2}} = \frac{a^{2} (x − \sqrt{x^{2} − a^{2}})^{2}}{[x^{2} − (x^{2} − a^{2})]^{2}} = \frac{(x − \sqrt{x^{2} − a^{2}})^{2}}{a^{2}}

= \frac{(x + \sqrt{x^{2} − a^{2}})^{2}}{8} − \frac{a^{2}}{2} ln |\frac{x + \sqrt{x^{2} − a^{2}}}{a} | − \frac{(x − \sqrt{x^{2} − a^{2}})^{2}}{8} + c =

= \frac{x^{2} + 2x\sqrt{x^{2} − a^{2}} + x^{2} − a^{2}}{8} − \frac{x^{2} − 2x\sqrt{x^{2} − a^{2}} + x^{2} − a^{2}}{8} − \frac{a^{2}}{2} ln |x + \sqrt{x^{2} − a^{2}}| + \frac{a^{2}}{2} ln a + c =

= \frac{4x\sqrt{x^{2} − a^{2}}}{8} − \frac{a^{2}}{2} ln |x + \sqrt{x^{2} − a^{2}}| + c = \frac{x\sqrt{x^{2} − a^{2}}}{2} − \frac{a^{2}}{2} ln |x + \sqrt{x^{2} − a^{2}}| + c, for a > 0, x > a.

∫ \sqrt{x^{2} − a^{2}} dx = ∫ \sqrt{x^{2} − (−a)^{2}} dx = \frac{x\sqrt{x^{2} − a^{2}}}{2} − \frac{a^{2}}{2} ln |x + \sqrt{x^{2} − a^{2}}| + c, for a < 0 [i.e. −a > 0, (−a)^{2} = a^{2}], x < a.

⟹ ∫ \sqrt{a^{2} + x^{2}} dx = \frac{x\sqrt{x^{2} − a^{2}}}{2} − \frac{a^{2}}{2} ln |x + \sqrt{x^{2} − a^{2}}| + c, for a∈R − \{0\}, x∈(−∞; − |a|) ∪ (|a|; ∞).

∫ \sqrt{x^{2} − a^{2}} dx = ∫ \frac{t^{2} + a^{2}}{2t} \frac{t^{2} + a^{2}}{2t^{2}} dt = ∫ \frac{t^{4} + 2t^{2}a^{2} + a^{4}}{4t^{3}} dt = \frac{1}{4} ∫ [t + 2a^{2}t^{−1} + a^{4}t^{−3}] dt =

1-st Euler: \sqrt{x^{2} − a^{2}} = t − x ⟹ x^{2} − a^{2} = t^{2} − 2tx + x^{2} ⟹ x = \frac{t^{2} + a^{2}}{2t} , t = x + \sqrt{x^{2} − a^{2}}

\sqrt{x^{2} − a^{2}} = t − \frac{t^{2} + a^{2}}{2t} = \frac{t^{2} − a^{2}}{2t} , dx = \frac{2t · 2t − 2(t^{2} − a^{2})}{4t^{2}} dt = \frac{2t^{2} + 2a^{2}}{4t^{2}} dt = \frac{t^{2} + a^{2}}{2t^{2}} dt

t^{2} = (x + \sqrt{x^{2} − a^{2}})^{2} , t^{−2} = \frac{1}{(x + \sqrt{x^{2} − a^{2}})^{2}} \begin{matrix}(x − \sqrt{x^{2} − a^{2}}) & 2 \\ (x − \sqrt{x^{2} − a^{2}}) & 2 \end{matrix} = \begin{matrix}(x − \sqrt{x^{2} − a^{2}}) & 2 \\ [x^{2} − (x^{2} − a^{2})] & 2 \end{matrix} = \frac{(x − \sqrt{x^{2} − a^{2}})^{2}}{a^{4}}

= \frac{1}{4} [\frac{t^{2}}{2} + 2a^{2} ln |t| + a^{4} \frac{t^{−2}}{−2}] + c = \frac{(x + \sqrt{x^{2} − a^{2}})^{2}}{8} − \frac{(x − \sqrt{x^{2} − a^{2}})^{2}}{8} + \frac{a^{2}}{2} ln |x + \sqrt{x^{2} − a^{2}}| + c =

= \frac{x^{2} + 2x\sqrt{x^{2} − a^{2}} + x^{2} − a^{2}}{8} − \frac{x^{2} − 2x\sqrt{x^{2} − a^{2}} + x^{2} − a^{2}}{8} + \frac{a^{2}}{2} ln |x + \sqrt{x^{2} − a^{2}}| + c =

= \frac{4x\sqrt{x^{2} − a^{2}}}{8} + \frac{a^{2}}{2} ln |x + \sqrt{x^{2} − a^{2}}| + c = \frac{x\sqrt{x^{2} − a^{2}}}{2} + \frac{a^{2}}{2} ln |x + \sqrt{x^{2} − a^{2}}| + c,

for a∈R − {0}, x∈(−∞; − |a|) ∪ (|a|; ∞).

∫ \sqrt{x^{2} − a^{2}} dx = ∫ \frac{x^{2} − a^{2}}{\sqrt{x^{2} − a^{2}}} dx = ∫ \frac{x^{2} dx}{\sqrt{x^{2} − a^{2}}} − ∫ \frac{a^{2} dx}{\sqrt{x^{2} − a^{2}}} = \begin{matrix}u = x ⇒ & u′ = 1 \\ v′ = \frac{x}{\sqrt{x^{2} − a^{2}}} ⇒ & v = \sqrt{x^{2} − a^{2}} \end{matrix} =

= [x\sqrt{x^{2} − a^{2}} − ∫ \sqrt{x^{2} − a^{2}} dx] − ∫ \frac{a^{2} dx}{\sqrt{x^{2} − a^{2}}} = x\sqrt{x^{2} − a^{2}} − ∫ \sqrt{x^{2} − a^{2}} dx − a^{2} ln |x + \sqrt{x^{2} − a^{2}}|

⟹^{(i.e.equation)} x\sqrt{x^{2} − a^{2}} − a^{2} ln |x + \sqrt{x^{2} − a^{2}}| + c_{0} = ∫ \sqrt{x^{2} − a^{2}} dx + ∫ \sqrt{x^{2} − a^{2}} dx = 2 ∫ \sqrt{x^{2} − a^{2}} dx

⟹ ∫ \sqrt{x^{2} − a^{2}} dx = \frac{x\sqrt{x^{2} − a^{2}}}{2} − \frac{a^{2}}{2} ln |x + \sqrt{x^{2} − a^{2}}| + c, for a∈R − \{0\}, x∈(−∞; − |a|) ∪ (|a|; ∞).

∫ arctg x dx = \begin{matrix}u′ = 1 ⇒ & u = x \\ v = arctg x ⇒ & v′ = \frac{1}{1 + x^{2}} \end{matrix} = x arctg x − ∫ \frac{x}{1 + x^{2}} dx = x arctg x − \frac{1}{2} ∫ \frac{2x}{1 + x^{2}} dx = x arctg x − \frac{1}{2} ln |1 + x^{2}| + c =

= x arctg x − \frac{1}{2} ln (1 + x^{2}) + c = x arctg x − ln \sqrt{1 + x^{2}} + c, for x∈R.

I_{n} =: ∫ x^{n} e^{x} dx = \begin{matrix}u = x^{n} ⇒ & u′ = nx^{n−1} \\ v′ = e^{x} ⇒ & v = e^{x} \end{matrix} = x^{n} e^{x} − ∫ nx^{n−1} e^{x} dx = x^{n} e^{x} − n ∫ x^{n−1} e^{x} dx = x^{n} e^{x} − nI_{n−1}, for n∈N, x∈R.

I_{0} = ∫ x^{0} e^{x} dx = ∫ e^{x} dx = e^{x} + c,

I_{1} = x^{1} e^{x} −1I_{0} = x e^{x} −1 e^{x} + c, I_{2} = x^{2} e^{x} −2I_{1} = x^{2} e^{x} −2 (x e^{x} −1 e^{x}) + c = x^{2} e^{x} −2x e^{x} +2 · 1 e^{x} + c, I_{3} = x^{3} e^{x} −3I_{2} = x^{3} e^{x} −3 (x^{2} e^{x} −2x e^{x} +2 · 1 e^{x}) + c = x^{3} e^{x} −3x^{2} e^{x} +3 · 2x e^{x} −3 · 2 · 1 e^{x} + c,

· · · · · · · · ·

I_{n} = x^{n} e^{x} −nI_{n−1} = ∑_{j=0}^{n}(−1)^{j}n(n − 1) ⋯ (n − j + 1)x^{n−j} e^{x} = e^{x} ∑_{j=0}^{n}(−1)^{j}n(n − 1) ⋯ (n − j + 1)x^{n−j} = e^{x} ∑_{j=0}^{n} \frac{(−1)^{j}x^{n−j}n!}{(n − j)!} =

= e^{x} [x^{n} − nx^{n−1} + n(n − 1)x^{n−2} − ⋯ + (−1)^{j}n(n − 1) ⋯ (n − j + 1)x^{n−j} + ⋯ + (−1)^{n}n!] + c, for n∈N, x∈R.

∫ \frac{dx}{\sqrt{x + 1} + \sqrt[3]{x + 1}} = ∫ \frac{dx}{(\sqrt[6]{x + 1})^{3} + (\sqrt[6]{x + 1})^{2}} = \begin{matrix}\sqrt[6]{x + 1} = t \\ x + 1 = t^{6} \\ dx = 6t^{5} dt \end{matrix} = ∫ \frac{6t^{5} dt}{t^{3} + t^{2}} = 6 ∫ \frac{t^{3} dt}{t + 1} = 6 ∫ \frac{t^{3} + t^{2} − t^{2} − t + t + 1 − 1}{t + 1} dt =

= 6 ∫ [\frac{t^{3} + t^{2}}{t + 1} − \frac{t^{2} + t}{t + 1} + \frac{t + 1}{t + 1} − \frac{1}{t + 1}] dt = 6 ∫ [t^{2} − t + 1 − \frac{1}{t + 1}] dt = 6 [\frac{t^{3}}{3} − \frac{t^{2}}{2} + t − ln |t + 1|] + c =

= 2 (\sqrt[6]{x + 1})^{3} − 3 (\sqrt[6]{x + 1})^{2} + 6\sqrt[6]{x + 1} − 6 ln |1 + \sqrt[6]{x + 1}| + c =

= 2\sqrt{x + 1} − 3\sqrt[3]{x + 1} + 6\sqrt[6]{x + 1} − 6 ln |1 + \sqrt[6]{x + 1}| + c, for x > −1.

∫ ln (1 + x^{2}) dx = \begin{matrix}u′ = 1 ⇒ & u = x \\ v = ln (1 + x^{2}) ⇒ & v′ = \frac{2x}{1 + x^{2}} \end{matrix} = x ln (1 + x^{2}) − ∫ \frac{2x^{2} dx}{1 + x^{2}} = x ln (1 + x^{2}) − 2 ∫ \frac{x^{2} + 1 − 1}{1 + x^{2}} dx =

= x ln (1 + x^{2}) − 2 ∫ [1 − \frac{1}{1 + x^{2}} ] dx = x ln (1 + x^{2}) − 2x + 2 arctg x + c, for x∈R.

∫ \frac{dx}{\sqrt{x^{2} + a^{2}}} = ∫ \frac{2t}{t^{2} + a^{2}} \frac{t^{2} + a^{2}}{2t^{2}} dt = ∫ \frac{dt}{t} = ln |t| + c = ln |x + \sqrt{x^{2} + a^{2}}| + c = ln (x + \sqrt{x^{2} + a^{2}}) + c, for a∈R − \{0\}, x∈R.

1-st Euler: \sqrt{x^{2} + a^{2}} = t − x ⟹ x^{2} + a^{2} = t^{2} − 2tx + x^{2} ⟹ 2tx = t^{2} − a^{2} ⟹ x = \frac{t^{2} − a^{2}}{2t}

\sqrt{x^{2} + a^{2}} = t − \frac{t^{2} − a^{2}}{2t} = \frac{t^{2} + a^{2}}{2t} , dx = \frac{2t · 2t − 2(t^{2} − a^{2})}{4t^{2}} dt = \frac{2t^{2} + 2a^{2}}{4t^{2}} dt = \frac{t^{2} + a^{2}}{2t^{2}} dt

\sqrt{a^{2} + x^{2}} > \sqrt{x^{2}} ≥ x ⟹ \sqrt{a^{2} + x^{2}} − x > 0 ⟹ ln |x + \sqrt{a^{2} + x^{2}}| = ln (x + \sqrt{a^{2} + x^{2}})

∫ \frac{dx}{\sqrt{x^{2} + a^{2}}} = ∫ \frac{cos t}{a} \frac{a dt}{cos^{2} t} = ∫ \frac{cos t dt}{1 − sin^{2} t} = u = sin t ; du = cos t dt = ∫ \frac{du}{1 − u^{2}} = − ∫ \frac{du}{u^{2} − 1} = −\frac{1}{2} ln |\frac{u − 1}{u + 1}| + c_{0} =

x = a tg t ⟹ dx = \frac{a dt}{cos^{2} t} ⟹ \sqrt{x^{2} + a^{2}} = \sqrt{\frac{a^{2} sin^{2} t}{cos^{2} t} + a^{2}} = \sqrt{\frac{a^{2} sin^{2} t + a^{2} cos^{2} t}{cos^{2} t}} = \sqrt{\frac{a^{2}}{cos^{2} t}} = \frac{|a|}{cos t} = \frac{a}{cos t}

sgn sin t = sgn tg t ⟹ sin t = \frac{tg t}{\sqrt{tg^{2} t + 1}} = \frac{\frac{x}{a}}{\sqrt{\frac{x^{2}}{a^{2}} + 1}} = \frac{x}{\sqrt{x^{2} + a^{2}}} ⟹ sin t ± 1 = \frac{x ± \sqrt{x^{2} + a^{2}}}{\sqrt{x^{2} + a^{2}}}

= \frac{1}{2} ln |\frac{u + 1}{u − 1}| + c_{0} = \frac{1}{2} ln |\frac{sin t + 1}{sin t − 1}| + c_{0} = \frac{1}{2} ln |\frac{x + \sqrt{x^{2} + a^{2}}}{x − \sqrt{x^{2} + a^{2}}} | + c_{0} = \frac{1}{2} ln |\frac{x + \sqrt{x^{2} + a^{2}}}{x − \sqrt{x^{2} + a^{2}}} · \frac{x + \sqrt{x^{2} + a^{2}}}{x + \sqrt{x^{2} + a^{2}}} | + c_{0} =

= \frac{1}{2} ln |\frac{(x + \sqrt{x^{2} + a^{2}})^{2}}{x^{2} − (x^{2} + a^{2})} | + c_{0} = ln |x + \sqrt{x^{2} + a^{2}}| − ln a + c_{0} = ln (x + \sqrt{x^{2} + a^{2}}) + c, for a > 0, x∈R.

∫ \frac{dx}{\sqrt{x^{2} + a^{2}}} = ∫ \frac{dx}{\sqrt{x^{2} + (−a)^{2}}} = ln (x + \sqrt{x^{2} + a^{2}}) + c, for a < 0 [i.e. −a > 0, (−a)^{2} = a^{2}], x∈R.

⟹ ∫ \frac{dx}{\sqrt{x^{2} + a^{2}}} = ln (x + \sqrt{x^{2} + a^{2}}) + c, for a∈R − \{0\}, x∈R.

∫ \frac{dx}{\sqrt{x^{2} − a^{2}}} = ∫ \frac{2t}{t^{2} − a^{2}} \frac{t^{2} − a^{2}}{2t^{2}} dt = ∫ \frac{dt}{t} = ln |t| + c = ln |x + \sqrt{x^{2} − a^{2}}| + c, for a∈R − \{0\}, x∈(−∞; − |a|) ∪ (|a|; ∞).

1-st Euler: \sqrt{x^{2} − a^{2}} = t − x ⟹ x^{2} − a^{2} = t^{2} − 2tx + x^{2} ⟹ 2tx = t^{2} + a^{2} ⟹ x = \frac{t^{2} + a^{2}}{2t}

\sqrt{x^{2} − a^{2}} = t − \frac{t^{2} + a^{2}}{2t} = \frac{t^{2} − a^{2}}{2t} , dx = \frac{2t · 2t − 2(t^{2} + a^{2})}{4t^{2}} dt = \frac{2t^{2} − 2a^{2}}{4t^{2}} dt = \frac{t^{2} − a^{2}}{2t^{2}} dt

x∈R ⟹ t = arctg x∈(−\frac{π}{2} ; \frac{π}{2} ) ⟹ \sqrt{cos^{2} x} = |cos x| = cos x, tg^{2} t = \frac{sin^{2} t}{cos^{2} t} = \frac{sin^{2} t}{1 − sin^{2} t} ⟹ sin^{2} t = \frac{tg^{2} t}{tg^{2} t + 1}

∫ \frac{dx}{\sqrt{a^{2} − x^{2}}} = x = a sin t dx = a cos t dt = ∫ \frac{a cos t dt}{a cos t} = ∫ dt = t + c = arcsin \frac{x}{a} + c,  for a > 0, x∈(−a; a).

t = arcsin \frac{x}{a}, \sqrt{a^{2} − x^{2}} = \sqrt{a^{2} − a^{2} sin^{2} t} = a\sqrt{1 − sin^{2} t} = a\sqrt{cos^{2} t} = a |cos t| = a cos t, x∈(−a; a) ⇒ t∈(−\frac{π}{2} ; \frac{π}{2} )

∫ \frac{dx}{\sqrt{a^{2} − x^{2}}} = ∫ \frac{dx}{\sqrt{(−a)^{2} − x^{2}}} = arcsin \frac{x}{−a} + c, for a < 0 [i.e. −a > 0, (−a)^{2} = a^{2}], x∈(a; −a).

⟹ ∫ \frac{dx}{\sqrt{a^{2} − x^{2}}} = arcsin \frac{x}{|a|} + c, for a∈R − \{0\}, x∈(− |a|; |a|).

∫ \frac{dx}{\sqrt{a^{2} − x^{2}}} = ∫ \frac{t^{2} + 1}{a(1 − t^{2})} \frac{2a(t^{2} − 1)}{(t^{2} + 1)^{2}} dt = −2 ∫ \frac{dt}{t^{2} + 1} = −2 arctg t + c = −2 arctg \frac{\sqrt{a^{2} − x^{2}} − a}{x} + c,

for a∈R − {0}, x∈(− |a|; |a|) − {0}.

2-nd Euler: \sqrt{a^{2} − x^{2}} = xt + a ⟹ a^{2} − x^{2} = x^{2}t^{2} + 2txa + a^{2} ⟹ −2txa = x^{2}t^{2} + x^{2} ⟹^{x≠0} x = \frac{−2ta}{t^{2} + 1}

\sqrt{a^{2} − x^{2}} = \frac{−2t^{2}a}{t^{2} + 1} + a = \frac{a(1 − t^{2})}{t^{2} + 1},  dx = \frac{−2a(t^{2} + 1) + 2ta · 2t}{(t^{2} + 1)^{2}} dt = \frac{2a(t^{2} − 1)}{(t^{2} + 1)^{2}} dt,  t = \frac{\sqrt{a^{2} − x^{2}} − a}{x}

∫ \begin{matrix}√ & \overline{1 − \sqrt{x}} \\ √ & \overline{1 + \sqrt{x}} \end{matrix} dx = ∫ \begin{matrix}√ & \overline{1 − \sqrt{x}} \\ √ & \overline{1 + \sqrt{x}} \end{matrix} · \begin{matrix}√ & \overline{1 − \sqrt{x}} \\ √ & \overline{1 − \sqrt{x}} \end{matrix} dx = ∫ \frac{1 − \sqrt{x}}{\sqrt{1 − x}} dx = ∫ \frac{dx}{\sqrt{1 − x}} − ∫ \begin{matrix}√ & \overline{x} \\ √ & \overline{1 − x} \end{matrix} dx = ∫ (1 − x)^{− \frac{1}{2}} dx − ∫ \sqrt{\frac{x}{1 − x}} dx =

\begin{matrix}1 − x = u \\ dx = − du \end{matrix} resp.  \frac{x}{1 − x} = t^{2} ⟹ x = t^{2} − xt^{2} ⟹ x = \frac{t^{2}}{1 + t^{2}} ⟹ dx = \frac{2t(1 + t^{2}) − t^{2} · 2t}{(1 + t^{2})^{2}} dt = \frac{2t dt}{(1 + t^{2})^{2}}

= − ∫ u^{− \frac{1}{2}} du − ∫ \frac{t · 2t dt}{(1 + t^{2})^{2}} = −\frac{u^{\frac{1}{2}}}{\frac{1}{2}} − 2 ∫ \frac{t^{2} dt}{(1 + t^{2})^{2}} = −2\sqrt{u} − 2 ∫ \frac{1 + t^{2} − 1}{(1 + t^{2})^{2}} dt =

= −2\sqrt{1 − x} − 2 ∫ [\frac{1}{1 + t^{2}} − \frac{1}{(1 + t^{2})^{2}} ] dt = −2\sqrt{1 − x} − 2 [arctg t − \frac{1}{2} arctg t − \frac{1}{2} \frac{t}{t^{2} + 1}] + c =

t^{2} + 1 = \frac{x}{1 − x} + 1 = \frac{x + 1 − x}{1 − x} = \frac{1}{1 − x} ⟹ \frac{t}{t^{2} + 1} = (1 − x)\sqrt{\frac{x}{1 − x}} = \sqrt{(1 − x)x} = \sqrt{x − x^{2}}

= −2\sqrt{1 − x} − arctg t + \frac{t}{t^{2} + 1} + c = −2\sqrt{1 − x} − arctg \sqrt{\frac{x}{1 − x}} + \sqrt{x − x^{2}} + c, for x∈⟨0 ; 1).

∫ |x| dx = \begin{cases}∫ x dx = \frac{x^{2}}{2} + c = \frac{x |x|}{2} + c,  for & x≥0, \\ − ∫ x dx = −\frac{x^{2}}{2} + c = \frac{x |x|}{2} + c, for & x<0. \end{cases}  ⟹ ∫ |x| dx = \frac{x |x|}{2} + c,  for x∈R.

∫ x^{2} e^{2x} dx = \begin{matrix}u = x^{2} ⇒ & u′ = 2x \\ v′ = e^{2x} ⇒ & v = \frac{e^{2x}}{2} \end{matrix} = \frac{x^{2} e^{2x}}{2} − ∫ \frac{2x e^{2x}}{2} dx = \frac{x^{2} e^{2x}}{2} − ∫ x e^{2x} dx = \begin{matrix}u = x ⇒ & u′ = 1 \\ v′ = e^{2x} ⇒ & v = \frac{e^{2x}}{2} \end{matrix} =

= \frac{x^{2} e^{2x}}{2} − [\frac{x e^{2x}}{2} −∫ \frac{e^{2x}}{2} dx] = \frac{x^{2} e^{2x}}{2} − \frac{x e^{2x}}{2} +∫ \frac{e^{2x}}{2} dx = \frac{x^{2} e^{2x}}{2} − \frac{x e^{2x}}{2} + \frac{e^{2x}}{4} + c = e^{2x} [\frac{x^{2}}{2} − \frac{x}{2} + \frac{1}{4}] + c, for x∈R.

∫ x^{2} e^{2x} dx = e^{2x} [Ax^{2} + Bx + C] + c = e^{2x} [\frac{x^{2}}{2} − \frac{x}{2} + \frac{1}{4}] + c, for x∈R.

[∫ x^{2} e^{2x} dx]′ = [e^{2x} (Ax^{2} + Bx + C) + c]′ ⟹ x^{2} e^{2x} = 2 e^{2x} (Ax^{2} + Bx + C) + e^{2x} (2Ax + B) ⟹

x^{2} = 2Ax^{2} + 2Bx + 2Ax + 2C + B ⟹ 1 = 2A, 0 = 2A + 2B, 0 = B + 2C ⟹ A = \frac{1}{2}, B = −\frac{1}{2}, C = \frac{1}{4}

∫ x^{9} e^{2x} dx = \begin{matrix}u = x^{9} ⇒ & u′ = 9x^{8} \\ v′ = e^{2x} ⇒ & v = \frac{e^{2x}}{2} \end{matrix} = \frac{x^{9} e^{2x}}{2} − ∫ \frac{9x^{8} e^{2x}}{2} dx = \begin{matrix}u = x^{8} ⇒ & u′ = 8x^{7} \\ v′ = e^{2x} ⇒ & v = \frac{e^{2x}}{2} \end{matrix} = · · ·. ^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}^{.}^{.}^{.}^{.}^{.}_{.}^{.}^{.}^{.}^{.}^{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.} _{.}^{.}^{.}^{.}^{.}^{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}^{.}^{.}^{.}^{.}^{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.} _{.}^{.}^{.}^{.}^{.}^{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.} _{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}^{^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}} ^{^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}}^{^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}} ^{^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}}^{^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}} ^{^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}}^{^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}} ^{^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}}_{. . . . . . . . . . . . . . . .}_{. . . . . . . . . . . . . . . .}

∫ x^{9} e^{2x} dx = e^{2x} [Ax^{9} + Bx^{8} + Cx^{7} + Dx^{6} + Ex^{5} + F x^{4} + Gx^{3} + Hx^{2} + Ix + J] + c

x^{9} e^{2x} = 2 e^{2x} (Ax^{9} + Bx^{8} + Cx^{7} + Dx^{6} + Ex^{5} + F x^{4} + Gx^{3} + Hx^{2} + Ix + J) + + e^{2x} (9Ax^{8} + 8Bx^{7} + 7Cx^{6} + 6Dx^{5} + 5Ex^{4} + 4F x^{3} + 3Gx^{2} + 2Hx + I) ⟹

x^{9} = 2Ax^{9} + (2B + 9A)x^{8} + (2C + 8B)x^{7} + (2D + 7C)x^{6} + (2E + 6D)x^{5} + (2F + 5E)x^{4}+ +(2G + 4F)x^{3} + (2H + 3G)x^{2} + (2I + 2H)x + 2J + I ⟹

A = \frac{1}{2}, B = −\frac{9}{4}, C = 9, D = −\frac{63}{2}, E = \frac{189}{2}, F = −\frac{945}{4}, G = \frac{945}{2}, H = −\frac{2835}{4}, I = \frac{2835}{4}, J = −\frac{2835}{8}

⟹ ∫ x^{9} e^{2x} dx = e^{2x} [\frac{x^{9}}{2} − \frac{9x^{8}}{4} + 9x^{7} − \frac{63x^{6}}{2} + \frac{189x^{5}}{2} − \frac{945x^{4}}{4} + \frac{945x^{3}}{2} − \frac{2835x^{2}}{4} + \frac{2835x}{4} − \frac{2835}{8}] + c, for x∈R.

∫ \sqrt{\frac{1 + x}{1 − x}} dx = \begin{matrix}t = \sqrt{\frac{1 + x}{1 − x}}, t^{2} = \frac{1 + x}{1 − x} ⟹ 1 + x = t^{2} − xt^{2} ⟹ x = \frac{t^{2} − 1}{1 + t^{2}}, x∈⟨−1 ; 1) \\ t^{2} + 1 = \frac{2}{1 − x}, dx = \frac{2t(1 + t^{2}) − 2t(t^{2} − 1)}{(1 + t^{2})^{2}} dt = \frac{4t dt}{(1 + t^{2})^{2}} , t∈⟨0 ; ∞) \end{matrix} = ∫ \frac{4t^{2} dt}{(1 + t^{2})^{2}} = 4 ∫ \frac{t^{2} + 1 − 1 dt}{(1 + t^{2})^{2}} =

= 4 ∫ [\frac{1}{1 + t^{2}} 1\frac{1}{(1 + t^{2})^{2}} ] dt = 4 [arctg t − \frac{1}{2} arctg t − \frac{1}{2} \frac{t}{t^{2} + 1}] + c = 2 arctg t − \frac{2t}{t^{2} + 1} + c =

= 2 arctg \sqrt{\frac{1 + x}{1 − x}} − 2\sqrt{\frac{1 + x}{1 − x}} \frac{1 − x}{2} + c = 2 arctg \sqrt{\frac{1 + x}{1 − x}} − \sqrt{1 − x^{2}} + c, for x∈⟨−1 ; 1).

∫ \sqrt{\frac{1 + x}{1 − x}} dx = \begin{matrix}x = −t \\ dx = − dt \end{matrix} = − ∫ \sqrt{(\frac{1 − t}{1 + t})^{3}} dt = − [\sqrt{1 − t^{2}} − 2 arctg \sqrt{\frac{1 − t}{1 + t}}] + c = 2 arctg \sqrt{\frac{1 + x}{1 − x}} − \sqrt{1 − x^{2}} + c,

for x∈⟨−1 ; 1).

∫ \frac{dx}{x^{2}\sqrt{x^{2} − 1}} = ∫ [\frac{2t}{t^{2} + 1}]^{2} \frac{2t}{t^{2} − 1} \frac{t^{2} − 1}{2t^{2}} dt = ∫ \frac{4t dt}{(t^{2} + 1)^{2}} = \begin{matrix}t^{2} + 1 = u \\ 2t dt = du \end{matrix} = ∫ \frac{2 du}{u^{2}} = 2 ∫ u^{−2} du =

1-st Euler: \sqrt{x^{2} − 1} = t − x ⟹ x^{2} − 1 = t^{2} − 2tx + x^{2} ⟹ 2tx = t^{2} + 1 ⟹ x = \frac{t^{2} + 1}{2t}

\sqrt{x^{2} − 1} = t − \frac{t^{2} + 1}{2t} = \frac{t^{2} − 1}{2t}, dx = \frac{2t · 2t − 2(t^{2} + 1)}{4t^{2}} dt = \frac{2t^{2} − 2}{4t^{2}} dt = \frac{t^{2} − 1}{2t^{2}} dt

= 2\frac{u^{−1}}{−1} + c_{1} = c_{1} − \frac{2}{t^{2} + 1} = c_{1} − \frac{2}{2tx} = c_{1} − \frac{1}{x (x + \sqrt{x^{2} − 1})}, for x∈(−∞; −1) ∪ (1 ; ∞).

= c_{1} − \frac{1}{x (x + \sqrt{x^{2} − 1})} · \frac{x − \sqrt{x^{2} − 1}}{x − \sqrt{x^{2} − 1}} = c_{1} − \frac{x − \sqrt{x^{2} − 1}}{x (x^{2} − x^{2} + 1)} = c_{1} − \frac{x − \sqrt{x^{2} − 1}}{x} = c_{1} − 1 + \frac{\sqrt{x^{2} − 1}}{x} = c_{2} + \frac{\sqrt{x^{2} − 1}}{x},

for x∈(−∞; −1) ∪ (1 ; ∞).

∫ \frac{dx}{x^{2}\sqrt{x^{2} − 1}} = \begin{matrix}x = \frac{1}{t}, t = \frac{1}{x} ⟹ dx = (t^{−1})′ dt = −\frac{dt}{t^{2}} \\ \sqrt{x^{2} − 1} = \sqrt{\frac{1}{t^{2}} − 1} = \frac{\sqrt{1 − t^{2}}}{t} for t∈(0 ; 1) \end{matrix} = − ∫ t^{2} \frac{t}{\sqrt{1 − t^{2}}} \frac{dt}{t^{2}} = ∫ \frac{−t dt}{\sqrt{1 − t^{2}}} = \begin{matrix}1 − t^{2} = u \\ −2t dt = du \end{matrix} =

= \frac{1}{2} ∫ \frac{du}{u^{\frac{1}{2}}} = \frac{1}{2} ∫ u^{− \frac{1}{2}} du = \frac{1}{2} \frac{u^{\frac{1}{2}}}{\frac{1}{2}} + c_{2} = \sqrt{1 − t^{2}} + c = \sqrt{1 − \frac{1}{x^{2}}} + c_{2} = \frac{\sqrt{x^{2} − 1}}{x} + c_{2}, for x∈(1 ; ∞).

∫ x^{x}(ln x + 1) dx = u = x^{x} ⇒ u′ = [e^{ln x^{x}}]′ = [e^{x ln x}]′ = e^{x ln x} [ln x + \frac{x}{x}] = x^{x}(ln x + 1) v′ = 1 ⇒ v = x = x^{x} + c, for x > 0.

∫ ln (x + \sqrt{x^{2} + 1}) dx = \begin{matrix}u = ln (x + \sqrt{x^{2} + 1}) ⇒ & u′ = \frac{1}{\sqrt{x^{2} + 1}} \\ v′ = 1 ⇒ & v = x \end{matrix} = x ln (x + \sqrt{x^{2} + 1}) − ∫ \frac{x dx}{\sqrt{x^{2} + 1}} = \begin{matrix}x^{2} + 1 = t \\ 2x dx = dt \end{matrix} =

= x ln (x + \sqrt{x^{2} + 1}) − \frac{1}{2} ∫ \frac{dt}{\sqrt{t}} = x ln (x + \sqrt{x^{2} + 1}) − \frac{1}{2} ∫ t^{− \frac{1}{2}} dt =

= x ln (x + \sqrt{x^{2} + 1}) − \frac{1}{2} \frac{t^{\frac{1}{2}}}{\frac{1}{2}} + c = x ln (x + \sqrt{x^{2} + 1}) − \sqrt{x^{2} + 1} + c, for x∈R.

∫ \frac{dx}{\sqrt{x − 3} + \sqrt{x − 4}} = ∫ \frac{1}{\sqrt{x − 3} + \sqrt{x − 4}} \frac{\sqrt{x − 3} − \sqrt{x − 4}}{\sqrt{x − 3} − \sqrt{x − 4}} dx = ∫ \frac{\sqrt{x − 3} − \sqrt{x − 4}}{x − 3 − (x − 4)} dx = ∫ [(x − 3)^{\frac{1}{2}} − (x − 4)^{\frac{1}{2}} ] dx =

= \frac{(x − 3)^{\frac{3}{2}}}{\frac{3}{2}} − \frac{(x − 4)^{\frac{3}{2}}}{\frac{3}{2}} + c = \frac{2\sqrt{(x − 3)^{3}}}{3} − \frac{2\sqrt{(x − 4)^{3}}}{3} + c, for x > 3.

∫ min_{x ∈ (0 ; ∞)} \{1, \frac{1}{x}\} dx = \begin{cases}∫ dx = x + c_{1}, for & x∈(0 ; 1⟩ , \\ ∫ \frac{dx}{x} = ln x + c_{2}, for & x∈⟨1 ; ∞) . \end{cases}

∫ \frac{2x^{2} − x + 1}{x\sqrt{1+x−x^{2}}} dx = ∫ \frac{2x − 1}{\sqrt{1 + x − x^{2}}} dx + ∫ \frac{dx}{x\sqrt{1 + x − x^{2}}} = ∫ \frac{− ds}{\sqrt{s}} − ∫ t\frac{t}{\sqrt{t^{2} + t − 1}} \frac{dt}{t^{2}} = − ∫ s^{− \frac{1}{2}} ds − ∫ \frac{dt}{\sqrt{(t + \frac{1}{2})^{2} − \frac{5}{4}}} =

\begin{matrix}1 + x − x^{2} = s \\ (1 − 2x) dx = ds \end{matrix}  \begin{matrix}x = \frac{1}{t}, t = \frac{1}{x} ⟹ dx = −\frac{dt}{t^{2}} ⟹ \sqrt{1 + x − x^{2}} = \sqrt{1 + \frac{1}{t} − \frac{1}{t^{2}}} = \frac{\sqrt{t^{2} + t − 1}}{t} for t∉⟨\frac{−1 − \sqrt{5}}{2} ; \frac{−1 + \sqrt{5}}{2}⟩ \\ x∈(\frac{1 − \sqrt{5}}{2} ; \frac{1 + \sqrt{5}}{2}) − \{0\} ⟹ \{ \begin{matrix}0 < x < \frac{1 + \sqrt{5}}{2} ⟹ t = \frac{1}{x} > \frac{2}{1 + \sqrt{5}} = \frac{2(1 − \sqrt{5})}{1 − 5} = \frac{\sqrt{5} − 1}{2} > 0 \\ \frac{1 − \sqrt{5}}{2} < x < 0 ⟹ t = \frac{1}{x} < \frac{2}{1 − \sqrt{5}} = \frac{2(1 + \sqrt{5})}{1 − 5} = −\frac{\sqrt{5} + 1}{2} < 0 \end{matrix} \\ t^{2} + t − 1 = (t + \frac{1}{2})^{2} − 1 − \frac{1}{4} = (t + \frac{1}{2})^{2} − \frac{5}{4} > 0 ⟹ |t + \frac{1}{2}| > \frac{\sqrt{5}}{2} ⟹ t + \frac{1}{2} < −\frac{\sqrt{5}}{2} or \frac{\sqrt{5}}{2} < t + \frac{1}{2} \end{matrix}

= \begin{matrix}t + 1/2 = u \\ dt = du \end{matrix} = −\frac{s^{\frac{1}{2}}}{\frac{1}{2}} − ∫ \frac{du}{\sqrt{u^{2} − \frac{5}{4}}} = −2\sqrt{s} − ln |u + \sqrt{u^{2} − \frac{5}{4}}| + c = −2\sqrt{1 + x − x^{2}} − ln |t+ \frac{1}{2} + \sqrt{t^{2} + t − 1}| + c =

= −2\sqrt{1 + x − x^{2}} − ln |\frac{1}{x} + \frac{1}{2} + \sqrt{\frac{1}{x^{2}} + \frac{1}{x} − 1}| + c = −2\sqrt{1 + x − x^{2}} − ln |\frac{2 + x}{2x} + \frac{\sqrt{1 + x − x^{2}}}{x} | + c =

= −2\sqrt{1 + x − x^{2}} − ln |\frac{2 + x + 2\sqrt{1 + x − x^{2}}}{2x} | + c, for x∈(\frac{1 − \sqrt{5}}{2} ; \frac{1 + \sqrt{5}}{2}) − \{0\}.

∫ \frac{2x^{2} − x + 1}{x\sqrt{1 + x − x^{2}}} dx = ∫ \frac{t^{4} + 2t^{3} + 9t^{2} − 6t + 2}{(t^{2} + 1)^{2}} · \frac{t^{2} + 1}{1 − 2t} · \frac{t^{2} + 1}{−t^{2} + t + 1} · \frac{2t^{2} − 2t − 2}{(t^{2} + 1)^{2}} dt = 2 ∫ \frac{t^{4} + 2t^{3} + 9t^{2} − 6t + 2}{(2t − 1)(t^{2} + 1)^{2}} dt =

⟹ x = \frac{1 − 2t}{t^{2} + 1},  \sqrt{1 + x − x^{2}} = xt + 1 = \frac{1 − 2t}{t^{2} + 1}t + 1 = \frac{t − 2t^{2} + t^{2} + 1}{t^{2} + 1} = \frac{−t^{2} + t + 1}{t^{2} + 1},  t = \frac{\sqrt{1 + x − x^{2}} − 1}{x}

dx = \frac{−2(t^{2} + 1) − 2t(1 − 2t)}{(t^{2} + 1)^{2}} dt = \frac{−2t^{2} − 2 − 2t + 4t^{2}}{(t^{2} + 1)^{2}} dt = \frac{2t^{2} − 2t − 2}{(t^{2} + 1)^{2}} dt, t^{2} = \frac{−x^{2} + x + 2 − 2\sqrt{1 + x − x^{2}}}{x^{2}}

2x^{2} − x + 1 = 2 [\frac{1 − 2t}{t^{2} + 1}]^{2} − \frac{1 − 2t}{t^{2} + 1} + 1 = \frac{2(1 − 4t + 4t^{2})}{(t^{2} + 1)^{2}} − \frac{t^{2} + 1 − 2t^{3} − 2t}{(t^{2} + 1)^{2}} + \frac{t^{4} + 2t^{2} + 1}{(t^{2} + 1)^{2}} = \frac{t^{4} + 2t^{3} + 9t^{2} − 6t + 2}{(t^{2} + 1)^{2}}

4 + 2t = \frac{4x + 2\sqrt{1 + x − x^{2}} − 2}{x}, t^{2} + 1 = \frac{−x^{2} + x + 2 − 2\sqrt{1 + x − x^{2}} + x^{2}}{x^{2}} = \frac{x + 2 − 2\sqrt{1 + x − x^{2}}}{x^{2}}

= 2 ∫ [\frac{1}{2t − 1} + \frac{1}{t^{2} + 1} + \frac{4t − 2}{(t^{2} + 1)^{2}} ] dt = ∫ \frac{2 dt}{2t − 1} + 2 ∫ \frac{dt}{t^{2} + 1} + 4 ∫ \frac{2t}{(t^{2} + 1)^{2}} dt − 4 ∫ \frac{1}{(t^{2} + 1)^{2}} dt =

\frac{t^{4} + 2t^{3} + 9t^{2} − 6t + 2}{(2t − 1)(t^{2} + 1)^{2}} = \frac{A}{2t − 1} + \frac{Bt + C}{t^{2} + 1} + \frac{Dt + E}{(t^{2} + 1)^{2}} ⟹ A = 1, B = 0, C = 1, D = 4, E = −2  \begin{matrix}2t−1 = u \\ 2 dt = du \end{matrix}  \begin{matrix}t^{2} + 1 = v \\ 2t dt = dv \end{matrix}

= ∫ \frac{du}{u} + 2 ∫ \frac{dt}{t^{2} + 1} + 4 ∫ \frac{dv}{v^{2}} − 4 ∫ \frac{dt}{(t^{2} + 1)^{2}} =_{17}^{page} ln |u| + 2 arctg t + 4\frac{v^{−1}}{−1} − 4 [\frac{1}{2} arctg t + \frac{1}{2} \frac{t}{t^{2} + 1}] + c =

1 + x − x^{2} > 0 ⟹ x^{2} − x − 1 = (x − \frac{1}{2})^{2} − \frac{1}{4} − 1 = (x − \frac{1}{2})^{2} − \frac{5}{4} < 0 ⟹ (x − \frac{1}{2})^{2} < \frac{5}{4} ⟹

|x − \frac{1}{2}| < \frac{\sqrt{5}}{2} ⟹ −\frac{\sqrt{5}}{2} < x − \frac{1}{2} < \frac{\sqrt{5}}{2} ⟹ \frac{1}{2} − \frac{\sqrt{5}}{2} < x < \frac{1}{2} + \frac{\sqrt{5}}{2} ⟹ x∈(\frac{1 − \sqrt{5}}{2} ; \frac{1 + \sqrt{5}}{2})

= ln |2t − 1| + 2 arctg t − \frac{4}{v} − 2 arctg t − \frac{2t}{t^{2} + 1} + c = ln |2t − 1| − \frac{4}{t^{2} + 1} − \frac{2t}{t^{2} + 1} + c = ln |2t − 1| − \frac{4 + 2t}{t^{2} + 1} + c =

= ln |\frac{2\sqrt{1 + x − x^{2}} − 2 − x}{x}| − x\frac{4x − 2 + 2\sqrt{1 + x − x^{2}}}{x + 2 − 2\sqrt{1 + x − x^{2}}} + c,  for x∈(\frac{1 − \sqrt{5}}{2} ; \frac{1 + \sqrt{5}}{2}) − \{0\}.  ^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}^{.}^{.}^{.}^{.}^{.}_{.}^{.}^{.}^{.}^{.}^{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.} _{.}^{.}^{.}^{.}^{.}^{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}^{.}^{.}^{.}^{.}^{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.} _{.}^{.}^{.}^{.}^{.}^{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.} _{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}_{.}^{^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}} ^{^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}}^{^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}} ^{^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}}^{^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}} ^{^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}}^{^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}} ^{^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}^{.}}_{. . . . . . . . . . . . . . . .}_{. . . . . . . . . . . . . . . .}

2-nd Euler: \sqrt{−x^{2} + x + 1} = xt + 1 ⟹ −x^{2} + x + 1 = x^{2}t^{2} + 2tx + 1 ⟹ x − 2tx = x^{2}t^{2} + x^{2} ⟹ 1 − 2t = xt^{2} + x

∫ \frac{\sqrt{x^{2} − x}}{x} dx = ∫ \begin{matrix}√ & \overline{x^{2} − x} \\ √ & \overline{x^{2}} \end{matrix} dx = ∫ \sqrt{\frac{x^{2} − x}{x^{2}}} dx = ∫ \sqrt{\frac{x − 1}{x}} dx = ∫ \frac{t · 2t dt}{(1 − t^{2})^{2}} = ∫ \frac{2t^{2} dt}{(1 − t^{2})^{2}} =

x ≠ 0, x^{2} − x = x(x − 1) ≥ 0 ⟹ x ≠ 0, x∈(−∞; 0⟩ ∪ ⟨1 ; ∞) ⟹ x∈(−∞; 0) ∪ ⟨1 ; ∞)

x ≥ 0: \sqrt{\frac{x − 1}{x}} = t, t∈(0 ; 1) ⟹ \frac{x − 1}{x} = t^{2} ⟹ x = \frac{1}{1 − t^{2}} ⟹ dx = \frac{−(−2t) dt}{(1 − t^{2})^{2}} = \frac{2t dt}{(1 − t^{2})^{2}}

1 ± t = 1 ± \begin{matrix}√ & \overline{x − 1} \\ √ & \overline{x} \end{matrix} = \frac{\sqrt{x} ± \sqrt{x − 1}}{\sqrt{x}}, \frac{t}{1 − t^{2}} = xt = x\sqrt{\frac{x − 1}{x}} = \sqrt{x}\sqrt{x − 1} = \sqrt{x^{2} − x}

\frac{1 − t}{1 + t} = \frac{\sqrt{x} − \sqrt{x − 1}}{\sqrt{x} + \sqrt{x − 1}} = \frac{\sqrt{x} − \sqrt{x − 1}}{\sqrt{x} + \sqrt{x − 1}} \frac{\sqrt{x} − \sqrt{x − 1}}{\sqrt{x} − \sqrt{x − 1}} = \frac{x − 2\sqrt{x(x − 1)} + x − 1}{x − (x − 1)} = 2x − 1 − 2\sqrt{x^{2} − x} > 0

= \frac{1}{2} ∫ [\frac{1}{t − 1} + \frac{1}{(t − 1)^{2}} − \frac{1}{t + 1} + \frac{1}{(t + 1)^{2}} ] dt = \frac{1}{2} [ln |t − 1| + \frac{(t − 1)^{−1}}{−1} − ln |t + 1| + \frac{(t + 1)^{−1}}{−1} ] + c =

\frac{2t^{2}}{(1 − t^{2})^{2}} = \frac{A}{t − 1} + \frac{B}{(t − 1)^{2}} + \frac{C}{t + 1} + \frac{D}{(t + 1)^{2}} ⟹ A = \frac{1}{2}, B = \frac{1}{2}, C = −\frac{1}{2}, D = \frac{1}{2}

= \frac{1}{2} [ln \frac{1 − t}{1 + t} + \frac{1}{1 − t} − \frac{1}{1 + t}] + c = \frac{1}{2} [ln \frac{1 − t}{1 + t} + \frac{1 + t − (1 − t)}{1 − t^{2}}] + c = \frac{1}{2} ln \frac{1 − t}{1 + t} + \frac{t}{1 − t^{2}} + c =

= \frac{1}{2} ln (2x − 1 − 2\sqrt{x^{2} − x}) + \sqrt{x^{2} − x} + c, for x∈⟨1 ; ∞).

∫ \frac{\sqrt{x^{2} − x}}{x} dx = ∫ \frac{\sqrt{x^{2} − x}}{−\sqrt{x^{2}}} dx = − ∫ \sqrt{\frac{x − 1}{x}} dx = − ∫ \frac{2t^{2} dt}{(1 − t^{2})^{2}} = −\frac{1}{2} ln \frac{1 − t}{1 + t} − \frac{t}{1 − t^{2}} + c =

x < 0: \sqrt{\frac{x − 1}{x}} = t, t∈(0 ; 1) ⟹ \frac{x − 1}{x} = t^{2} ⟹ x = \frac{1}{1 − t^{2}} ⟹ dx = \frac{2t dt}{(1 − t^{2})^{2}}

1 ± t = 1 ± \begin{matrix}√ & \overline{1 − x} \\ √ & \overline{−x} \end{matrix} = \frac{\sqrt{−x} ± \sqrt{1 − x}}{\sqrt{−x}}, \frac{t}{1 − t^{2}} = xt = −\sqrt{(−x)^{2}}\sqrt{\frac{1 − x}{−x}} = −\sqrt{−x}\sqrt{1 − x} = −\sqrt{x^{2} − x}

\frac{1 + t}{1 − t} = \frac{\sqrt{−x} + \sqrt{1 − x}}{\sqrt{−x} − \sqrt{1 − x}} = \frac{\sqrt{−x} + \sqrt{1 − x}}{\sqrt{−x} − \sqrt{1 − x}} \frac{\sqrt{−x} + \sqrt{1 − x}}{\sqrt{−x} + \sqrt{1 − x}} = \frac{−x + 2\sqrt{−x(1 − x)} + 1 − x}{−x − (1 − x)} = 2x − 1 − 2\sqrt{x^{2} − x} > 0

= \frac{1}{2} ln \frac{1 + t}{1 − t} − \frac{t}{1 − t^{2}} + c = \frac{1}{2} ln (2x − 1 − 2\sqrt{x^{2} − x}) + \sqrt{x^{2} − x} + c, for x∈(−∞; 0).

⟹ ∫ \frac{\sqrt{x^{2} − x}}{x} dx = \frac{1}{2} ln (2x − 1 − 2\sqrt{x^{2} − x}) + \sqrt{x^{2} − x} + c, for x∈(−∞; 0) ∪ ⟨1 ; ∞).

∫ \frac{x^{5} dx}{\sqrt{x^{3} + 1}} = \frac{1}{3} ∫ \frac{x^{3} · 3x^{2} dx}{\sqrt{x^{3} + 1}} = \begin{matrix}x^{3} + 1 = t \\ 3x^{2} dx = dt \end{matrix} = \frac{1}{3} ∫ \frac{(t − 1) dt}{\sqrt{t}} = \frac{1}{3} ∫ [\sqrt{t} − \frac{1}{\sqrt{t}}] dt = \frac{1}{3} ∫ [t^{\frac{1}{2}} − t^{− \frac{1}{2}} ] dt = \frac{1}{3} [\frac{t^{\frac{3}{2}}}{\frac{3}{2}} − \frac{t^{\frac{1}{2}}}{\frac{1}{2}} ] + c =

= \frac{2\sqrt{t^{3}}}{9} − \frac{2\sqrt{t}}{3} + c = \frac{2\sqrt{(x^{3} + 1)^{3}}}{9} − \frac{2\sqrt{x^{3} + 1}}{3} + c, for x∈(−1 ; ∞).

∫ \frac{dx}{\sqrt{5 + 4 e^{x}}} = \begin{matrix}e^{x} = t, t > 0 \\ x = ln t, dx = \frac{dt}{t} \end{matrix} = ∫ \frac{dt}{t\sqrt{5 + 4t}} = \begin{matrix}4t = u^{2} \\ 4 dt = 2u du \end{matrix} = ∫ \frac{4}{u^{2}\sqrt{5 + u^{2}}} \frac{u du}{2} = 2 ∫ \frac{du}{u\sqrt{5 + u^{2}}} =

= \begin{matrix}u = \frac{\sqrt{5}}{v} ⟹ du = −\frac{\sqrt{5} dv}{v^{2}} , u > 0 \\ \sqrt{5 + u^{2}} = \sqrt{5 + \frac{5}{v^{2}}} = \frac{\sqrt{5}}{v}\sqrt{v^{2} + 1}, v > 0 \end{matrix} = 2 ∫ \frac{v}{\sqrt{5}} \frac{v}{\sqrt{5}\sqrt{v^{2} + 1}} \frac{−\sqrt{5} dv}{v^{2}} = −\frac{2}{\sqrt{5}} ∫ \frac{dv}{\sqrt{v^{2} + 1}} =

= −\frac{2}{\sqrt{5}} ln (v + \sqrt{v^{2} + 1}) + c_{1} = −\frac{2}{\sqrt{5}} ln (\frac{\sqrt{5}}{u} + \sqrt{\frac{5}{u^{2}} + 1}) + c_{1} = −\frac{2}{\sqrt{5}} ln (\frac{\sqrt{5}}{u} + \frac{\sqrt{5 + u^{2}}}{u} ) + c_{1} =

= −\frac{2}{\sqrt{5}} ln \frac{\sqrt{5} + \sqrt{5 + u^{2}}}{u} + c_{1} = \frac{2}{\sqrt{5}} [ln u − ln (\sqrt{5} + \sqrt{5 + u^{2}})] + c_{1} = \frac{2}{\sqrt{5}} [ln (4t)^{\frac{1}{2}} − ln (\sqrt{5} + \sqrt{5 + 4t})] + c_{1} =

= \frac{2}{\sqrt{5}} [\frac{1}{2} ln (4 e^{x}) − ln (\sqrt{5} + \sqrt{5 + 4 e^{x}})] + c_{1} = \frac{2}{\sqrt{5}} [\frac{1}{2} ln 4 + \frac{1}{2} ln e^{x} − ln (\sqrt{5} + \sqrt{5 + 4 e^{x}})] + c_{1} =

= \frac{ln 4}{\sqrt{5}} + \frac{x}{\sqrt{5}} − \frac{2}{\sqrt{5}} ln (\sqrt{5} + \sqrt{5 + 4 e^{x}}) + c_{1} = \frac{x}{\sqrt{5}} − \frac{2}{\sqrt{5}} ln (\sqrt{5} + \sqrt{5 + 4 e^{x}}) + c, for x∈R.

∫ \frac{dx}{\sqrt{5 + 4 e^{x}}} = e^{x} = t, 4t = u^{2}, u = \frac{\sqrt{5}}{v} ⟹ e^{x} = t = \frac{u^{2}}{4} = \frac{5}{4v^{2}} , v^{2} = \frac{5}{4 e^{x}} v = \frac{\sqrt{5}}{2\sqrt{e^{x}}} , x = ln \frac{5}{4v^{2}} = ln 5 − ln 4 − 2 ln v ⟹ dx = \frac{−2 dv}{v} = ∫ \frac{1}{\sqrt{5 + 4\frac{5}{4v^{2}}}} \frac{−2 dv}{v} =

= −\frac{2}{\sqrt{5}} ∫ \frac{dv}{\sqrt{v^{2} + 1}} = −\frac{2}{\sqrt{5}} ln (v + \sqrt{v^{2} + 1}) + c_{1} = −\frac{2}{\sqrt{5}} ln (\frac{\sqrt{5}}{2\sqrt{e^{x}}} + \sqrt{\frac{5}{4 e^{x}} + 1}) + c_{1} =

= −\frac{2}{\sqrt{5}} ln \frac{\sqrt{5} + \sqrt{5 + 4 e^{x}}}{2\sqrt{e^{x}}} + c_{1} = −\frac{2}{\sqrt{5}} ln \frac{\sqrt{5} + \sqrt{5 + 4 e^{x}}}{2 (e^{x})^{\frac{1}{2}}} + c_{1} = −\frac{2}{\sqrt{5}} [ln (\sqrt{5} + \sqrt{5 + 4 e^{x}}) − ln 2 − \frac{1}{2} ln e^{x}] + c_{1} =

= \frac{2 ln 2}{\sqrt{5}} + \frac{x}{\sqrt{5}} − \frac{2}{\sqrt{5}} ln (\sqrt{5} + \sqrt{5 + 4 e^{x}}) + c_{1} = \frac{x}{\sqrt{5}} − \frac{2}{\sqrt{5}} ln (\sqrt{5} + \sqrt{5 + 4 e^{x}}) + c, for x∈R.

∫ \frac{\sqrt{1 − x^{2}}}{x^{2}} dx = \begin{matrix}u = \sqrt{1 − x^{2}} ⇒ & u′ = \frac{1}{2} (1 − x^{2})^{− \frac{1}{2}} (−2x) = \frac{−x}{\sqrt{1 − x^{2}}} \\ v′ = \frac{1}{x^{2}} = x^{−2} ⇒ & v = \frac{x^{−1}}{−1} = −\frac{1}{x} \end{matrix} = −\frac{\sqrt{1 − x^{2}}}{x} − ∫ \frac{−x}{\sqrt{1 − x^{2}}} \frac{− dx}{x} =

= −\frac{\sqrt{1 − x^{2}}}{x} − ∫ \frac{dx}{\sqrt{1 − x^{2}}} = −\frac{\sqrt{1 − x^{2}}}{x} − arcsin x + c, for x∈⟨−1 ; 1⟩ − \{0\}.

∫ \frac{\sqrt{1 − x^{2}}}{x^{2}} dx = x = sin t, t = arcsin x ⟹ dx = cos t dt,  x∈⟨−1 ; 1⟩ \sqrt{1 − x^{2}} = \sqrt{1 − sin^{2} t} = \sqrt{cos^{2} t} = |cos t| = cos t, t∈⟨−\frac{π}{2} ; \frac{π}{2} ⟩ = ∫ \frac{cos t · cos t dt}{sin^{2} t} = ∫ \frac{1 − sin^{2} t dt}{sin^{2} t} =

= ∫ [\frac{1}{sin^{2} t} − 1] dt = − cotg t − t + c = c − \frac{cos t}{sin t} − t = c − \frac{\sqrt{1 − x^{2}}}{x} − arcsin x, for x∈⟨−1 ; 1⟩ − \{0\}.

∫ \sqrt{(\frac{1 − x}{1 + x})^{3}} dx = \begin{matrix}t = \sqrt{\frac{1 − x}{1 + x}}, t^{2} = \frac{1 − x}{1 + x} ⟹ 1 − x = t^{2} + xt^{2} ⟹ x = \frac{1 − t^{2}}{1 + t^{2}} , x∈(−1 ; 1⟩ \\ t^{2} + 1 = \frac{2}{1 + x}, dx = \frac{−2t(1 + t^{2}) − 2t(1 − t^{2})}{(1 + t^{2})^{2}} dt = \frac{−4t dt}{(1 + t^{2})^{2}} , t∈⟨0 ; ∞) \end{matrix} = ∫ \frac{−4t^{4} dt}{(1 + t^{2})^{2}} =

\frac{t^{4}}{(1 + t^{2})^{2}} = \frac{t^{4} + 2t^{2} + 1 − 2t^{2} − 2 + 1}{(1 + t^{2})^{2}} = \frac{t^{4} + 2t^{2} + 1}{(1 + t^{2})^{2}} − \frac{2(t^{2} + 1)}{(1 + t^{2})^{2}} + \frac{1}{(1 + t^{2})^{2}} = 1 − \frac{2}{1 + t^{2}} + \frac{1}{(1 + t^{2})^{2}}

= −4 ∫ [1 − \frac{2}{1 + t^{2}} + \frac{1}{(1 + t^{2})^{2}} ] dt =_{17}^{page} −4 [t − 2 arctg t + \frac{1}{2} arctg t + \frac{1}{2} \frac{t}{t^{2} + 1}] + c = 6 arctg t − 4t − \frac{2t}{t^{2} + 1} + c =

= 6 arctg \sqrt{\frac{1 − x}{1 + x}} − 4\sqrt{\frac{1 − x}{1 + x}} − 2\sqrt{\frac{1 − x}{1 + x}} \frac{1 + x}{2} + c = 6 arctg \sqrt{\frac{1 − x}{1 + x}} − 4\sqrt{\frac{1 − x}{1 + x}} − \sqrt{1 − x^{2}} + c, for x∈(−1 ; 1⟩.

∫ \sqrt{(\frac{1 + x}{1 − x})^{3}} dx = \begin{matrix}t = \sqrt{\frac{1 + x}{1 − x}}, t^{2} = \frac{1 + x}{1 − x} ⟹ 1 + x = t^{2} − xt^{2} ⟹ x = \frac{t^{2} − 1}{1 + t^{2}}, x∈⟨−1 ; 1) \\ t^{2} + 1 = \frac{2}{1 − x}, dx = \frac{2t(1 + t^{2}) − 2t(t^{2} − 1)}{(1 + t^{2})^{2}} dt = \frac{4t dt}{(1 + t^{2})^{2}} , t∈⟨0 ; ∞) \end{matrix} = ∫ \frac{4t^{4} dt}{(1 + t^{2})^{2}} =

\frac{t^{4}}{(1 + t^{2})^{2}} = \frac{t^{4} + 2t^{2} + 1 − 2t^{2} − 2 + 1}{(1 + t^{2})^{2}} = \frac{t^{4} + 2t^{2} + 1}{(1 + t^{2})^{2}} − \frac{2(t^{2} + 1)}{(1 + t^{2})^{2}} + \frac{1}{(1 + t^{2})^{2}} = 1 − \frac{2}{1 + t^{2}} + \frac{1}{(1 + t^{2})^{2}}

= 4 ∫ [1 − \frac{2}{1 + t^{2}} + \frac{1}{(1 + t^{2})^{2}} ] dt = 4 [t − 2 arctg t + \frac{1}{2} arctg t + \frac{1}{2} \frac{t}{t^{2} + 1}] + c = 4t − 6 arctg t + \frac{2t}{t^{2} + 1} + c =

= 4\sqrt{\frac{1 + x}{1 − x}} − 6 arctg \sqrt{\frac{1 + x}{1 − x}} + 2\sqrt{\frac{1 + x}{1 − x}} \frac{1 − x}{2} + c = 4\sqrt{\frac{1 + x}{1 − x}} − 6 arctg \sqrt{\frac{1 + x}{1 − x}} + \sqrt{1 − x^{2}} + c, for x∈⟨−1 ; 1).

∫ \sqrt{(\frac{1 + x}{1 − x})^{3}} dx = \begin{matrix}x = −t \\ dx = − dt \end{matrix} = − ∫ \sqrt{(\frac{1 − t}{1 + t})^{3}} dt = − [6 arctg \sqrt{\frac{1 − t}{1 + t}} − 4\sqrt{\frac{1 − t}{1 + t}} − \sqrt{1 − t^{2}}] + c =

= −6 arctg \sqrt{\frac{1 + x}{1 − x}} + 4\sqrt{\frac{1 + x}{1 − x}} + \sqrt{1 − x^{2}} + c, for x∈⟨−1 ; 1).

∫ x^{2} ln \sqrt{1 − x} dx = ∫ x^{2} ln (1 − x)^{\frac{1}{2}} dx = ∫ \frac{x^{2}}{2} ln (1 − x) dx = \begin{matrix}u = ln (1 − x) ⇒ & u′ = \frac{−1}{1 − x} = \frac{1}{x − 1} \\ v′ = \frac{x^{2}}{2} ⇒ & v = \frac{x^{3}}{2 · 3} = \frac{x^{3}}{6} \end{matrix} =

= \frac{x^{3} ln (1 − x)}{6} − ∫ \frac{x^{3}}{6} \frac{dx}{x − 1} = \frac{x^{3} ln (1 − x)}{6} − \frac{1}{6} ∫ \frac{(x^{3} − x^{2}) + (x^{2} − x) + (x − 1) + 1}{x − 1} dx =

= \frac{x^{3} ln (1 − x)}{6} − \frac{1}{6} ∫ [x^{2} + x + 1 + \frac{1}{x − 1}] dx = \frac{x^{3} ln (1 − x)}{6} − \frac{1}{6} [\frac{x^{2}}{3} + \frac{x^{2}}{2} + x + ln |x − 1|] + c =

= \frac{x^{3} ln (1 − x)}{6} − \frac{x^{3}}{18} − \frac{x^{2}}{12} − \frac{x}{6} − \frac{ln (1 − x)}{6} + c = \frac{x^{3} − 1}{6} ln (1 − x) − \frac{x^{3}}{18} − \frac{x^{2}}{12} − \frac{x}{6} + c, for x < −1.

∫ \frac{dx}{x^{6}(1 + x^{2})} = ∫ [\frac{1}{x^{2}} − \frac{1}{x^{4}} + \frac{1}{x^{6}} − \frac{1}{1 + x^{2}} ] dx = ∫ [x^{−2} − x^{−4} + x^{−6} − \frac{1}{1 + x^{2}} ] dx = \frac{x^{−1}}{−1} − \frac{x^{−3}}{−3} + \frac{x^{−5}}{−5} − arctg x + c =

x^{2} = t ⟹ \frac{1}{x^{6}(1 + x^{2})} = \frac{1}{t^{3}(1 + t)} = \frac{A}{t} + \frac{B}{t^{2}} + \frac{C}{t^{3}} + \frac{D}{1 + t} = \frac{A}{x^{2}} + \frac{B}{x^{4}} + \frac{C}{x^{6}} + \frac{D}{1 + x^{2}} ⟹ A = 1, B = −1, C = 1, D = −1

= −\frac{1}{x} + \frac{1}{3x^{3}} − \frac{1}{5x^{5}} − arctg x + c, for x∈R − \{0\}.

∫ \sqrt{\frac{1 − x}{1 + x}} dx = \begin{matrix}t = \sqrt{\frac{1 − x}{1 + x}}, t^{2} = \frac{1 − x}{1 + x} ⟹ 1 − x = t^{2} + xt^{2} ⟹ x = \frac{1 − t^{2}}{1 + t^{2}} , x∈(−1 ; 1⟩ \\ t^{2} + 1 = \frac{1 − x}{1 + x} + 1 = \frac{2}{1 + x}, dx = \frac{−2t(1 + t^{2}) − 2t(1 − t^{2})}{(1 + t^{2})^{2}} dt = \frac{−4t dt}{(1 + t^{2})^{2}} , t∈⟨0 ; ∞) \end{matrix} = ∫ \frac{−4t^{2} dt}{(1 + t^{2})^{2}} =

= −4 ∫ \frac{t^{2} + 1 − 1}{(1 + t^{2})^{2}} dt = −4 ∫ [\frac{1}{1 + t^{2}} − \frac{1}{(1 + t^{2})^{2}} ] dt = −4 [arctg t − \frac{1}{2} arctg t − \frac{1}{2} \frac{t}{t^{2} + 1}] + c =_{17}^{page}

= −2 arctg t + \frac{2t}{t^{2} + 1} + c = 2\sqrt{\frac{1 − x}{1 + x}} \frac{1 + x}{2} − 2 arctg \sqrt{\frac{1 − x}{1 + x}} + c = \sqrt{1 − x^{2}} − 2 arctg \sqrt{\frac{1 − x}{1 + x}} + c, for x∈(−1 ; 1⟩.

∫ \sqrt{\frac{1 + x}{1 − x}} dx = \begin{matrix}t = \sqrt{\frac{1 + x}{1 − x}}, t^{2} = \frac{1 + x}{1 − x} ⟹ 1 + x = t^{2} − xt^{2} ⟹ x = \frac{t^{2} − 1}{1 + t^{2}}, x∈⟨−1 ; 1) \\ t^{2} + 1 = \frac{1 + x}{1 − x} + 1 = \frac{2}{1 − x}, dx = \frac{2t(1 + t^{2}) − 2t(t^{2} − 1)}{(1 + t^{2})^{2}} dt = \frac{4t dt}{(1 + t^{2})^{2}} , t∈⟨0 ; ∞) \end{matrix} = ∫ \frac{4t^{2} dt}{(1 + t^{2})^{2}} =

= 4 ∫ \frac{t^{2} + 1 − 1 dt}{(1 + t^{2})^{2}} = 4 ∫ [\frac{1}{1 + t^{2}} 1\frac{1}{(1 + t^{2})^{2}} ] dt = 4 [arctg t − \frac{1}{2} arctg t − \frac{1}{2} \frac{t}{t^{2} + 1}] + c = 2 arctg t − \frac{2t}{t^{2} + 1} + c =

= 2 arctg \sqrt{\frac{1 + x}{1 − x}} − 2\sqrt{\frac{1 + x}{1 − x}} \frac{1 − x}{2} + c = 2 arctg \sqrt{\frac{1 + x}{1 − x}} − \sqrt{1 − x^{2}} + c, for x∈⟨−1 ; 1).

∫ \sqrt{\frac{1 + x}{1 − x}} dx = \begin{matrix}x = −t \\ dx = − dt \end{matrix} = −∫ \sqrt{\frac{1 − t}{1 + t}} dt = − [\sqrt{1 − t^{2}} − 2 arctg \sqrt{\frac{1 − t}{1 + t}}] + c = 2 arctg \sqrt{\frac{1 + x}{1 − x}} − \sqrt{1 − x^{2}} + c, for x∈⟨−1 ; 1).

∫ \sqrt{\frac{x − 1}{x + 1}} dx = ∫ \frac{4t^{2} dt}{(t − 1)^{2} (t + 1)^{2}} = \frac{4t^{2}}{(t^{2} − 1)^{2}} = \frac{A}{t − 1} + \frac{B}{(t − 1)^{2}} + \frac{C}{t + 1} + \frac{D}{(t + 1)^{2}} ⟹ A = 1, B = 1, C = −1, D = 1 =

t = \sqrt{\frac{x − 1}{x + 1}}, t^{2} = \frac{x − 1}{x + 1} ⟹ x − 1 = xt^{2} + t^{2} ⟹ x = \frac{1 + t^{2}}{1 − t^{2}} , x∈(−∞; −1) ∪ ⟨1 ; ∞) , i.e. x∈R − ⟨−1 ; 1)

1 − t^{2} = 1 − \frac{x − 1}{x + 1} = \frac{2}{x + 1} ⟹ \frac{1}{t^{2} − 1} = −\frac{x + 1}{2}, dx = \frac{2t(1 − t^{2}) − (−2t)(1 + t^{2})}{(1 − t^{2})^{2}} dt = \frac{4t dt}{(t^{2} − 1)^{2}} , t∈⟨0 ; ∞) − \{1\}

= ∫ [\frac{1}{t − 1} + \frac{1}{(t − 1)^{2}} − \frac{1}{t + 1} + \frac{1}{(t + 1)^{2}} ] dt = ln |t − 1| + ∫ (t − 1)^{−2} dt − ln |t + 1| + ∫ (t + 1)^{−2} dt =

= ln |\frac{t − 1}{t + 1}| + \frac{(t − 1)^{−1}}{−1} + \frac{(t + 1)^{−1}}{−1} + c = ln |\frac{t − 1}{t + 1}| − \frac{1}{t − 1} − \frac{1}{t + 1} + c = ln |t − 1| − ln |t + 1| − \frac{2t}{t^{2} − 1} + c =

= ln |\sqrt{\frac{x − 1}{x + 1}} − 1| − ln |\sqrt{\frac{x − 1}{x + 1}} + 1| + (x + 1)\sqrt{\frac{x − 1}{x + 1}} + c, for x∈(−∞; −1) ∪ ⟨1 ; ∞).

∫ \sqrt{\frac{x + 1}{x − 1}} dx = ∫ \frac{−4t^{2} dt}{(t − 1)^{2} (t + 1)^{2}} = \frac{4t^{2}}{(t^{2} − 1)^{2}} = \frac{A}{t − 1} + \frac{B}{(t − 1)^{2}} + \frac{C}{t + 1} + \frac{D}{(t + 1)^{2}} ⟹ A = 1, B = 1, C = −1, D = 1 =

t = \sqrt{\frac{x + 1}{x − 1}}, t^{2} = \frac{x + 1}{x − 1} ⟹ x + 1 = xt^{2} − t^{2} ⟹ x = \frac{t^{2} + 1}{t^{2} − 1}, x∈(−∞; −1) ∪ ⟨1 ; ∞) , i.e. x∈R − ⟨−1 ; 1)

t^{2} − 1 = \frac{x + 1}{x − 1} − 1 = \frac{2}{x − 1} ⟹ \frac{1}{t^{2} − 1} = \frac{x − 1}{2}, dx = \frac{2t(t^{2} − 1) − 2t(t^{2} + 1)}{(t^{2} − 1)^{2}} dt = \frac{−4t dt}{(t^{2} − 1)^{2}} , t∈⟨0 ; ∞) − \{1\}

= − ∫ [\frac{1}{t − 1} + \frac{1}{(t − 1)^{2}} − \frac{1}{t + 1} + \frac{1}{(t + 1)^{2}} ] dt = − ln |t − 1| − ∫ (t − 1)^{−2} dt + ln |t + 1| − ∫ (t + 1)^{−2} dt =

= ln |\frac{t + 1}{t − 1}| − \frac{(t − 1)^{−1}}{−1} − \frac{(t + 1)^{−1}}{−1} + c = ln |\frac{t + 1}{t − 1}| + \frac{1}{t − 1} + \frac{1}{t + 1} + c = ln |t + 1| − ln |t − 1| + \frac{2t}{t^{2} − 1} + c =

= ln |\sqrt{\frac{x + 1}{x − 1}} + 1| − ln |\sqrt{\frac{x + 1}{x − 1}} − 1| + (x − 1)\sqrt{\frac{x + 1}{x − 1}} + c, for x∈(−∞; −1) ∪ ⟨1 ; ∞).

∫ \frac{dx}{x\sqrt{1 − x^{3} + x^{6}}} = x^{6} − x^{3} + 1 = [x^{3} − \frac{1}{2}]^{2} + 1 − \frac{1}{4} > 0 = \frac{1}{3} ∫ \frac{3x^{2} dx}{x^{3}\sqrt{1 − x^{3} + x^{6}}} = \begin{matrix}t = x^{3} \\ 3x^{2} dx = dt \end{matrix} = \frac{1}{3} ∫ \frac{dt}{t\sqrt{1 − t + t^{2}}} =

t = \frac{1}{u} ⟹ dt = −\frac{du}{u^{2}}, \sqrt{1 − t + t^{2}} = \sqrt{1 − \frac{1}{u} + \frac{1}{u^{2}}} = \frac{\sqrt{u^{2} − u + 1}}{u}, \sqrt{u^{2} − u + 1} = \sqrt{\frac{1}{t^{2}} − \frac{1}{t} + 1} = \frac{\sqrt{1 − t + t^{2}}}{t}

= \frac{1}{3} ∫ u\frac{u}{\sqrt{u^{2} − u + 1}} \frac{− du}{u^{2}} = −\frac{1}{3} ∫ \frac{du}{\sqrt{(u − \frac{1}{2})^{2} + 1 − \frac{1}{4}}} = \begin{matrix}v = u − 1/2 \\ dv = du \end{matrix} = −\frac{1}{3} ∫ \frac{dv}{\sqrt{v^{2} + \frac{3}{4}}} =

= −\frac{1}{3} ln |v + \sqrt{v^{2} + \frac{3}{4}}| + c_{1} = c_{1} − \frac{1}{3} ln |u − \frac{1}{2} + \sqrt{u^{2} − u + 1}| = c_{1} − \frac{1}{3} ln |\frac{1}{t} − \frac{1}{2} + \frac{\sqrt{1 − t + t^{2}}}{t} | =

= c_{1} − \frac{1}{3} ln |\frac{2 − t + 2\sqrt{1 − t + t^{2}}}{2t} | = c_{1} − \frac{1}{3} [ln |2 − t + 2\sqrt{1 − t + t^{2}}| − ln |t| − ln 2] =

= c_{1} − \frac{1}{3} ln |2 − x^{3} + 2\sqrt{1 − x^{3} + x^{6}}| + \frac{1}{3} ln |x^{3}| + \frac{1}{3} ln 2 = c − \frac{1}{3} ln |2 − x^{3} + 2\sqrt{1 − x^{3} + x^{6}}| + ln x, for x∈R − \{0\}.

∫ \frac{dx}{x\sqrt{1 + x^{3} + x^{6}}} = x^{6} + x^{3} + 1 = [x^{3} + \frac{1}{2}]^{2} + 1 − \frac{1}{4} > 0 = \frac{1}{3} ∫ \frac{3x^{2} dx}{x^{3}\sqrt{1 + x^{3} + x^{6}}} = \begin{matrix}t = x^{3} \\ 3x^{2} dx = dt \end{matrix} = \frac{1}{3} ∫ \frac{dt}{t\sqrt{1 + t + t^{2}}} =

t = \frac{1}{u} ⟹ dt = −\frac{du}{u^{2}}, \sqrt{1 + t + t^{2}} = \sqrt{1 + \frac{1}{u} + \frac{1}{u^{2}}} = \frac{\sqrt{u^{2} + u + 1}}{u}, \sqrt{u^{2} + u + 1} = \sqrt{\frac{1}{t^{2}} + \frac{1}{t} + 1} = \frac{\sqrt{1 + t + t^{2}}}{t}

= \frac{1}{3} ∫ u\frac{u}{\sqrt{u^{2} + u + 1}} \frac{− du}{u^{2}} = −\frac{1}{3} ∫ \frac{du}{\sqrt{(u + \frac{1}{2})^{2} + 1 − \frac{1}{4}}} = \begin{matrix}v = u + 1/2 \\ dv = du \end{matrix} = −\frac{1}{3} ∫ \frac{dv}{\sqrt{v^{2} + \frac{3}{4}}} =

= −\frac{1}{3} ln |v + \sqrt{v^{2} + \frac{3}{4}}| + c_{1} = c_{1} − \frac{1}{3} ln |u + \frac{1}{2} + \sqrt{u^{2} + u + 1}| = c_{1} − \frac{1}{3} ln |\frac{1}{t} + \frac{1}{2} + \frac{\sqrt{1 + t + t^{2}}}{t} | =

= c_{1} − \frac{1}{3} ln |\frac{2 + t + 2\sqrt{1 + t + t^{2}}}{2t} | = c_{1} − \frac{1}{3} [ln |2 + t + 2\sqrt{1 + t + t^{2}}| − ln |t| − ln 2] =

= c_{1} − \frac{1}{3} ln |2 + x^{3} + 2\sqrt{1 + x^{3} + x^{6}}| + \frac{1}{3} ln |x^{3}| + \frac{1}{3} ln 2 = c − \frac{1}{3} ln |2 + x^{3} + 2\sqrt{1 + x^{3} + x^{6}}| + ln x, for x∈R − \{0\}.

∫ ln (\sqrt{1 + x} + \sqrt{1 − x}) dx = x ln (\sqrt{1 + x} + \sqrt{1 − x})−∫ x [\frac{1}{2x} − \frac{1}{2x\sqrt{1 − x^{2}}} ] dx = x ln (\sqrt{1 + x} + \sqrt{1 − x})−∫ \frac{dx}{2} +∫ \frac{dx}{2\sqrt{1 − x^{2}}} =

v′ = 1  ⇒ v = x

u = ln (\sqrt{1 + x} + \sqrt{1 − x}) ⇒ u′ = \frac{\frac{1}{2} \frac{1}{\sqrt{1 + x}} − \frac{1}{2} \frac{1}{\sqrt{1 − x}}}{\sqrt{1 + x} + \sqrt{1 − x}} = \frac{\frac{\sqrt{1 − x} − \sqrt{1 + x}}{2\sqrt{1 + x}\sqrt{1 − x}}}{\sqrt{1 + x} + \sqrt{1 − x}} = \frac{\sqrt{1 − x} − \sqrt{1 + x}}{\sqrt{1 + x} + \sqrt{1 − x}} \frac{1}{2\sqrt{1 − x^{2}}} =

= \frac{\sqrt{1 − x} − \sqrt{1 + x}}{\sqrt{1 + x} + \sqrt{1 − x}} \frac{1}{2\sqrt{1 − x^{2}}} \frac{\sqrt{1 − x} − \sqrt{1 + x}}{\sqrt{1 − x} − \sqrt{1 + x}} = \frac{(\sqrt{1 − x} − \sqrt{1 + x})^{2}}{1 − x − (1 + x)} \frac{1}{2\sqrt{1 − x^{2}}} =

= \frac{1 − x − 2\sqrt{1 − x}\sqrt{1 + x} + 1 + x}{−2x · 2\sqrt{1 − x^{2}}} = \frac{2 − 2\sqrt{1 − x^{2}}}{−2x · 2\sqrt{1 − x^{2}}} = \frac{\sqrt{1 − x^{2}} − 1}{2x\sqrt{1 − x^{2}}} = \frac{1}{2x} − \frac{1}{2x\sqrt{1 − x^{2}}}

= x ln (\sqrt{1 + x} + \sqrt{1 − x}) − \frac{x}{2} + \frac{arcsin x}{2} + c, for x∈⟨−1 ; 1⟩.

∫ \frac{1 − x}{x − \sqrt{x − x^{2}}} dx = ∫ \frac{t^{2}}{1 + t^{2}} \frac{1 + t^{2}}{1 − t} \frac{−2t dt}{(1 + t^{2})^{2}} = ∫ \frac{2t^{3} dt}{(t − 1) (1 + t^{2})^{2}} = ∫ [\frac{\frac{1}{2}}{t − 1} + \frac{\frac{3}{2} − \frac{t}{2}}{1 + t^{2}} + \frac{t − 1}{(1 + t^{2})^{2}} ] dt =

3-rd Euler: t = \sqrt{\frac{1 − x}{x}} ⟹ t^{2} = \frac{1 − x}{x} ⟹ t^{2}x = 1 − x ⟹ x = \frac{1}{1 + t^{2}} ⟹ dx = \frac{−1 · 2t}{(1 + t^{2})^{2}} dt = \frac{−2t dt}{(1 + t^{2})^{2}}

x − x^{2} = x(1 − x) > 0 ⟹ x∈(0 ; 1) ⟹ t∈(0 ; ∞) , 1 − x = 1 − \frac{1}{1 + t^{2}} = \frac{1 + t^{2} − 1}{1 + t^{2}} = \frac{t^{2}}{1 + t^{2}}

x − \sqrt{x(1 − x)} = \frac{1}{1 + t^{2}} − \sqrt{\frac{1}{1 + t^{2}} \frac{t^{2}}{1 + t^{2}}} = \frac{1}{1 + t^{2}} − \frac{t}{1 + t^{2}} = \frac{1 − t}{1 + t^{2}} , t = \sqrt{\frac{1 − x}{x}} = \sqrt{\frac{x(1 − x)}{x^{2}}} = \frac{\sqrt{x − x^{2}}}{x}

\frac{t}{1 + t^{2}} = \frac{1}{1 + t^{2}} t = x\frac{\sqrt{x − x^{2}}}{x} = \sqrt{x − x^{2}}, \frac{1 + t}{1 + t^{2}} = \frac{1}{1 + t^{2}} + \frac{t}{1 + t^{2}} = x + \sqrt{x − x^{2}}

\frac{2t^{3}}{(t − 1) (1 + t^{2})^{2}} = \frac{A}{t − 1} + \frac{Bt + C}{1 + t^{2}} + \frac{Dt + E}{(1 + t^{2})^{2}} ⟹ A = \frac{1}{2}, B = −\frac{1}{2}, C = \frac{3}{2}, D = 1, E = −1

= \frac{1}{2} ∫ \frac{dt}{t − 1} + \frac{3}{2} ∫ \frac{dt}{1 + t^{2}} − \frac{1}{2} ∫ \frac{t dt}{1 + t^{2}} + ∫ \frac{t dt}{(1 + t^{2})^{2}} − ∫ \frac{dt}{(1 + t^{2})^{2}} = \begin{matrix}1 + t^{2} = u \\ 2t dt = du \end{matrix} =

= \frac{1}{2} ln |t − 1| + \frac{3}{2} arctg t − \frac{1}{4} ∫ \frac{du}{u} + \frac{1}{2} ∫ \frac{du}{u^{2}} − ∫ \frac{dt}{(1 + t^{2})^{2}} =_{17}^{page} ∫ \frac{dx}{(x^{2} + 1)^{2}} = \frac{1}{2} arctg x + \frac{1}{2} \frac{x}{x^{2} + 1} =

= \frac{1}{2} ln |t − 1|+\frac{3}{2} arctg t−\frac{1}{4} ln |u|+\frac{1}{2} \frac{u^{−1}}{−1} −\frac{1}{2} arctg t+\frac{t}{2(t^{2} + 1)}+c = \frac{1}{4} ln (t − 1)^{2}+arctg t−\frac{1}{4} ln u−\frac{1}{2u}+\frac{t}{2(t^{2} + 1)}+c =

= \frac{1}{4} ln (t − 1)^{2} + arctg t − \frac{1}{4} ln (1 + t^{2}) − \frac{1}{2(1 + t^{2})} − \frac{t}{2(1 + t^{2})} + c = \frac{1}{4} ln \frac{t^{2} − 2t + 1}{1 + t^{2}} + arctg t − \frac{1 + t}{2(1 + t^{2})} + c =

= \frac{1}{4} ln (1 − \frac{2t}{1 + t^{2}} ) + arctg t − \frac{1 + t}{2(1 + t^{2})} + c = \frac{1}{4} ln (1 − 2\sqrt{x − x^{2}}) + arctg \frac{\sqrt{x − x^{2}}}{x} − \frac{x + \sqrt{x − x^{2}}}{2} + c,

for x∈(0 ; 1).

∫ \frac{1 − x}{x − \sqrt{x^{2} − x}} dx = ∫ \frac{(1−x) (x + \sqrt{x^{2}−x})}{x^{2} − (x^{2} − x)} dx = ∫ \frac{x − x^{2} + \sqrt{x^{2}−x} − x\sqrt{x^{2}−x}}{x} dx = ∫ (1 − x) dx + ∫ \frac{\sqrt{x^{2}−x}}{x} dx − ∫ \sqrt{x^{2}−x} dx =

= x − \frac{x^{2}}{2} + ∫ \frac{\sqrt{x^{2}−x}}{x} dx − ∫ \sqrt{(x− \frac{1}{2})^{2}− \frac{1}{4}} dx = \begin{matrix}x − 1/2 = t \\ dx = dt \end{matrix} = x − \frac{x^{2}}{2} + ∫ \frac{\sqrt{x^{2}−x}}{x} dx − ∫ \sqrt{t^{2}− \frac{1}{4}} dt =_{49,56}^{page}

∫ \frac{\sqrt{x^{2} − x}}{x} dx = \frac{1}{2} ln (2x − 1 − 2\sqrt{x^{2} − x}) + \sqrt{x^{2} − x}, ∫ \sqrt{x^{2} − a^{2}} dx = \frac{x\sqrt{x^{2} − a^{2}}}{2} − \frac{a^{2}}{2} ln |x + \sqrt{x^{2} − a^{2}}|

= x − \frac{x^{2}}{2} + [\frac{1}{2} ln (2x − 1 − 2\sqrt{x^{2} − x}) + \sqrt{x^{2} − x}] − [\frac{t\sqrt{t^{2} − 1/4}}{2} − \frac{1}{2} · \frac{1}{4} ln |t + \sqrt{t^{2} − 1/4}|] + c_{1} =

= x − \frac{x^{2}}{2} + \frac{1}{2} ln (2x − 1 − 2\sqrt{x^{2} − x}) + \sqrt{x^{2} − x} − \frac{2x − 1}{4}\sqrt{x^{2} − x} + \frac{1}{8} ln |x − \frac{1}{2} + \sqrt{x^{2} − x}| + c_{1} =

= x − \frac{x^{2}}{2} + \frac{4}{8} ln (2x − 1 − 2\sqrt{x^{2} − x}) + \frac{5 − 2x}{4}\sqrt{x^{2} − x} + \frac{1}{8} ln |\frac{2x − 1 + 2\sqrt{x^{2} − x}}{2}| + c_{1} =

= x − \frac{x^{2}}{2} + \frac{1 + 3}{4} ln (2x − 1 − 2\sqrt{x^{2} − x}) + \frac{5 − 2x}{4}\sqrt{x^{2} − x} + \frac{1}{8} ln |2x − 1 + 2\sqrt{x^{2} − x}| − \frac{1}{8} ln 2 + c_{1} =

= x − \frac{x^{2}}{2} + \frac{1}{8} ln |(2x − 1 − 2\sqrt{x^{2} − x}) (2x − 1 + 2\sqrt{x^{2} − x})| + \frac{3}{8} ln (2x − 1 − 2\sqrt{x^{2} − x}) + \frac{5 − 2x}{4}\sqrt{x^{2} − x} + c =

= x − \frac{x^{2}}{2} + \frac{1}{8} ln |(2x − 1)^{2} − 4(x^{2} − x)| + \frac{3}{8} ln (2x − 1 − 2\sqrt{x^{2} − x}) + \frac{5 − 2x}{4}\sqrt{x^{2} − x} + c =

= x − \frac{x^{2}}{2} + \frac{1}{8} ln |4x^{2} − 4x + 1 − 4x^{2} + 4x| + \frac{3}{8} ln (2x − 1 − 2\sqrt{x^{2} − x}) + \frac{5 − 2x}{4}\sqrt{x^{2} − x} + c =

= x − \frac{x^{2}}{2} + \frac{1}{8} ln 1 + \frac{3}{8} ln (2x − 1 − 2\sqrt{x^{2} − x}) + \frac{5 − 2x}{4}\sqrt{x^{2} − x} + c =

= x − \frac{x^{2}}{2} + \frac{3}{8} ln (2x − 1 − 2\sqrt{x^{2} − x}) + \frac{5 − 2x}{4}\sqrt{x^{2} − x} + c, for x∈(−∞; 0) ∪ (1 ; ∞).

∫ \frac{1 + \sqrt{1 − x^{2}}}{1 − \sqrt{1 − x^{2}}} dx = ∫ \frac{1 + \sqrt{1 − x^{2}}}{1 − \sqrt{1 − x^{2}}} \frac{1 + \sqrt{1 − x^{2}}}{1 + \sqrt{1 − x^{2}}} dx = ∫ \frac{1 + 2\sqrt{1 − x^{2}} + 1 − x^{2}}{1 − (1 − x^{2})} dx = ∫ \frac{2 − x^{2} + 2\sqrt{1 − x^{2}}}{x^{2}} dx =

= 2 ∫ x^{−2} dx − ∫ dx + 2 ∫ \frac{\sqrt{1 − x^{2}}}{x^{2}} dx =_{57}^{page} 2\frac{x^{−1}}{−1} − x + 2 [−\frac{\sqrt{1 − x^{2}}}{x} − arcsin x] + c =

= −\frac{2}{x} − x − \frac{2\sqrt{1 − x^{2}}}{x} − 2 arcsin x + c = c − \frac{2 + 2\sqrt{1 − x^{2}}}{x} − 2 arcsin x − x, for x∈⟨−1 ; 1⟩ − \{0\}.

∫ \sqrt{\frac{x}{1 − x\sqrt{x}}} dx = \begin{matrix}\sqrt{x} = t, x = t^{2}, dx = 2t dt \\ x∈⟨0 ; 1) , t∈⟨0 ; 1) \end{matrix} = ∫ \sqrt{\frac{t^{2}}{1 − t^{3}}} 2t dt = ∫ \frac{2t^{2} dt}{\sqrt{1 − t^{3}}} = \begin{matrix}1 − t^{3} = u \\ −3t^{2} dt = du \end{matrix} = −\frac{2}{3} ∫ \frac{du}{\sqrt{u}} =

x ≥ 0, 1 − x\sqrt{x} > 0 ⟹ x ≥ 0, 1 > x\sqrt{x} = \sqrt{x^{3}} ⟹ x ≥ 0, 1 > x ⟹ x∈⟨0 ; 1)

= −\frac{2}{3} ∫ u^{− \frac{1}{2}} du = −\frac{2}{3} \frac{u^{\frac{1}{2}}}{\frac{1}{2}} + c = c − \frac{4}{3}\sqrt{u} = c − \frac{4}{3}\sqrt{1 − t^{3}} = c − \frac{4}{3}\sqrt{1 − x\sqrt{x}}, for x∈⟨0 ; 1).

∫ \frac{dx}{\sqrt{e^{2x} + e^{x} +1}} = e^{x} = t > 0, x = ln t ∈ (−∞; ∞) , dx = \frac{dt}{t} = ∫ \frac{dt}{t\sqrt{t^{2} + t + 1}} = − ∫ \frac{\frac{du}{u^{2}}}{\frac{1}{u} \frac{\sqrt{1+u+u^{2}}}{u}} = − ∫ \frac{du}{\sqrt{1 + u + u^{2}}} =

t = \frac{1}{u} ⟹ dt = −\frac{du}{u^{2}}, \sqrt{t^{2} + t + 1} = \sqrt{\frac{1}{u^{2}} + \frac{1}{u} + 1} = \frac{\sqrt{1 + u + u^{2}}}{u} , \sqrt{1 + u + u^{2}} = \sqrt{1 + \frac{1}{t} + \frac{1}{t^{2}}} = \frac{\sqrt{t^{2} + t + 1}}{t}

= − ∫ \frac{du}{\sqrt{(u + \frac{1}{2})^{2} + 1 − \frac{1}{4}}} = u + \frac{1}{2} = v, du = dv = − ∫ \frac{dv}{\sqrt{v^{2} + \frac{3}{4}}} = − ln [v + \sqrt{v^{2} + \frac{3}{4}}] + c_{1} =

= c_{1} − ln [u + \frac{1}{2} + \sqrt{1 + u + u^{2}}] = c_{1} − ln [\frac{1}{t} + \frac{1}{2} + \frac{\sqrt{t^{2} + t + 1}}{t}] = c_{1} − ln [\frac{2 + t + 2\sqrt{t^{2} + t + 1}}{2t}] =

= c_{1} − ln (2 + t + 2\sqrt{t^{2} + t + 1}) + ln 2 + ln t = c − ln (2 + e^{x} +2\sqrt{e^{2x} + e^{x} +1}) + x, for x∈R.

∫ \sqrt{\frac{1 − e^{x}}{1 + e^{x}}} dx = \begin{matrix}e^{x} = t ⟹ x = ln t, dx = \frac{dt}{t} \\ x∈(−∞; 0⟩ ⟹ t∈(0 ; 1⟩ \end{matrix} = ∫ \sqrt{\frac{1 − t}{1 + t}} \frac{dt}{t} = ∫ u\frac{1 + u^{2}}{1 − u^{2}} \frac{−4u du}{(1 + u^{2})^{2}} = ∫ \frac{−4u^{2} du}{(1 − u^{2})(1 + u^{2})} =

u = \sqrt{\frac{1 − t}{1 + t}}, u^{2} = \frac{1 − t}{1 + t} ⟹ u^{2} + u^{2}t = 1 − t, t = \frac{1 − u^{2}}{1 + u^{2}} , dt = \frac{−2u(1 + u^{2}) − 2u(1 − u^{2})}{(1 + u^{2})^{2}} du = \frac{−4u du}{(1 + u^{2})^{2}} , u∈⟨0 ; 1)

1 ± u = 1 ± \sqrt{\frac{1 − t}{1 + t}} = \frac{\sqrt{1 + t} ± \sqrt{1 − t}}{\sqrt{1 + t}} ⟹ \frac{1 − u}{1 + u} = \frac{\sqrt{1 + e^{x}} − \sqrt{1 − e^{x}}}{\sqrt{1 + e^{x}} + \sqrt{1 − e^{x}}} = \frac{1 + e^{x} −2\sqrt{1 − e^{2x}} + 1 − e^{x}}{(1 + e^{x}) − (1 − e^{x})} = \frac{1 − \sqrt{1 − e^{2x}}}{e^{x}}

\frac{−4u^{2}}{(1 − u^{2})(1 + u^{2})} = \frac{2}{1 + u^{2}} − \frac{2}{1 − u^{2}} = \frac{2}{1 + u^{2}} − \frac{2}{(u − 1)(u + 1)} = \frac{2}{1 + u^{2}} − [\frac{1}{u + 1} − \frac{1}{u − 1}] = \frac{2}{1 + u^{2}} − \frac{1}{u + 1} + \frac{1}{u − 1}

= ∫ [\frac{2}{1 + u^{2}} − \frac{1}{u + 1} + \frac{1}{u − 1}] du = 2 arctg u − ln |u + 1| + ln |u − 1| + c = 2 arctg u + ln |\frac{u − 1}{u + 1}| + c =

= 2 arctg u + ln \frac{1 − u}{1 + u} + c = 2 arctg \sqrt{\frac{1 − e^{x}}{1 + e^{x}}} + ln \frac{1 − \sqrt{1 − e^{2x}}}{e^{x}} + c =

= 2 arctg \sqrt{\frac{1 − e^{x}}{1 + e^{x}}} +ln (1 − \sqrt{1−e^{2x}})−ln e^{x} +c = 2 arctg \sqrt{\frac{1 − e^{x}}{1 + e^{x}}} + ln (1−\sqrt{1 − e^{2x}}) − x + c, for x∈(−∞; 0⟩.

∫ \frac{1 − x}{x\sqrt{x − x^{2}}} dx = ∫ \frac{t^{2}}{1 + t^{2}} \frac{1 + t^{2}}{1} \frac{1 + t^{2}}{t} \frac{−2t dt}{(1 + t^{2})^{2}} = −2 ∫ \frac{t^{2} dt}{1 + t^{2}} = −2 ∫ \frac{1 + t^{2} − 1}{1 + t^{2}} dt = −2 ∫ [1 − \frac{1}{1 + t^{2}} ] dt =

3-rd Euler: t = \sqrt{\frac{1 − x}{x}} ⟹ t^{2} = \frac{1 − x}{x} ⟹ x = \frac{1}{1 + t^{2}} , 1 − x = \frac{t^{2}}{1 + t^{2}} ⟹ dx = \frac{−1 · 2t}{(1 + t^{2})^{2}} dt = \frac{−2t dt}{(1 + t^{2})^{2}}

x − x^{2} = x(1 − x) > 0 ⟹ x∈(0 ; 1) ⟹ t∈(1 ; ∞) , \sqrt{x(1 − x)} = \sqrt{\frac{1}{1 + t^{2}} \frac{t^{2}}{1 + t^{2}}} = \frac{t}{1 + t^{2}}

= −2(t − arctg t) + c = 2 arctg \sqrt{\frac{1 − x}{x}} − 2\sqrt{\frac{1 − x}{x}} + c, for x∈(0 ; 1).

∫ \frac{dx}{a^{2} cos^{2} x + b^{2} sin^{2} x} = sin^{2} x = \frac{1 − cos 2x}{2}, cos^{2} x = \frac{1 + cos 2x}{2} = ∫ \frac{2 dx}{a^{2}(1 + cos 2x) + b^{2}(1 − cos 2x)} =

= ∫ \frac{2 dx}{(a^{2} + b^{2}) + (a^{2} − b^{2}) cos 2x} = \begin{matrix}2x = t \\ 2x dx = dt \end{matrix} = ∫ \frac{dt}{(a^{2} + b^{2}) + (a^{2} − b^{2}) cos t} = ∫ \frac{u^{2} + 1}{2b^{2}u^{2} + 2a^{2}} \frac{2 du}{u^{2} + 1} = ∫ \frac{du}{b^{2}u^{2} + a^{2}} =

u=tg \frac{t}{2}, dt= \frac{2 du}{u^{2}+1}, cos t= \frac{1−u^{2}}{u^{2}+1} ⟹ (a^{2}+b^{2}) + (a^{2}−b^{2}) cos t=(a^{2}+b^{2})\frac{u^{2}+1}{u^{2}+1} + (a^{2}−b^{2})\frac{1−u^{2}}{u^{2}+1} = \frac{2b^{2}u^{2}+2a^{2}}{u^{2}+1}

= \frac{1}{b^{2}} ∫ \frac{du}{u^{2} + \frac{a^{2}}{b^{2}}} = \frac{1}{b^{2}} \frac{b}{a} arctg \frac{bu}{a} + c = \frac{1}{ab} arctg (\frac{b}{a} tg \frac{t}{2}) + c = \frac{1}{ab} arctg (\frac{b}{a} tg x) + c, for x∈R, a, b∈R − \{0\}.

∫ \frac{dx}{a^{2} cos^{2} x − b^{2} sin^{2} x} = sin^{2} x = \frac{1 − cos 2x}{2}, cos^{2} x = \frac{1 + cos 2x}{2} = ∫ \frac{2 dx}{a^{2}(1 + cos 2x) − b^{2}(1 − cos 2x)} =

= ∫ \frac{2 dx}{(a^{2} − b^{2}) + (a^{2} + b^{2}) cos 2x} = \begin{matrix}2x = t \\ 2x dx = dt \end{matrix} = ∫ \frac{dt}{(a^{2} − b^{2}) + (a^{2} + b^{2}) cos t} = ∫ \frac{u^{2} + 1}{2a^{2} − 2b^{2}u^{2}} \frac{2 du}{u^{2} + 1} = ∫ \frac{du}{a^{2} − b^{2}u^{2}} =

u=tg \frac{t}{2}, dt= \frac{2 du}{u^{2}+1}, cos t= \frac{1−u^{2}}{u^{2}+1} ⟹ (a^{2}−b^{2}) + (a^{2}+b^{2}) cos t=(a^{2}−b^{2})\frac{u^{2}+1}{u^{2}+1} + (a^{2}+b^{2})\frac{1−u^{2}}{u^{2}+1} = \frac{2a^{2}−2b^{2}u^{2}}{u^{2}+1}

= −\frac{1}{b^{2}} ∫ \frac{du}{u^{2} − \frac{a^{2}}{b^{2}}} = −\frac{1}{b^{2}} \frac{b}{2a} ln |\frac{u − \frac{a}{b}}{u + \frac{a}{b}} | + c = −\frac{1}{2ab} ln |\frac{bu − a}{bu + a}| + c = −\frac{1}{2ab} ln |\frac{b tg x − a}{b tg x + a}| + c, for x∈R, a, b∈R − \{0\}.

∫ x tg^{2} x dx = ∫ \frac{x sin^{2} x}{cos^{2} x} dx = ∫ \frac{x(1 − cos^{2} x)}{cos^{2} x} dx = ∫ \frac{x dx}{cos^{2} x} − ∫ x dx = ∫ \frac{x dx}{cos^{2} x} − \frac{x^{2}}{2} = u = x ⇒ u′ = 1 v′ = \frac{1}{cos^{2} x} ⇒ v = tg x = \frac{sin x}{cos x} =

= x tg x − ∫ \frac{sin x}{cos x} dx − \frac{x^{2}}{2} = x tg x + ∫ \frac{− sin x}{cos x} dx − \frac{x^{2}}{2} = x tg x + ln |cos x| − \frac{x^{2}}{2} + c, for x∈R − \{(2k + 1)\frac{π}{2} , k ∈Z\}.

∫ \frac{dx}{1 + \sqrt[3]{x}} = \begin{matrix}x ≥ 0, x = t^{3} ≥ 0 \\ t = \sqrt[3]{x}, dx = 3t^{2} dt \end{matrix} = ∫ \frac{3t^{2} dt}{1 + t} = 3 ∫ \frac{t^{2} dt}{1 + t} = 3 ∫ \frac{t^{2} + t − t − 1 + 1}{1 + t} dt = 3 ∫ [t − 1 + \frac{1}{1 + t}] dt =

= \frac{3t^{2}}{2} − 3t + 3 ln |1 + t| + c = \frac{3t^{2}}{2} − 3t + 3 ln (1 + t) + c = \frac{3\sqrt[3]{x^{2}}}{2} − 3\sqrt[3]{x} + 3 ln (1 + \sqrt[3]{x}) + c, for x∈⟨0 ; ∞)

∫ \frac{dx}{\sqrt[3]{x} + \sqrt[4]{x}} = ∫ \frac{12t^{11} dt}{t^{4} + t^{3}} = 12 ∫ \frac{t^{8} dt}{t + 1} = 12 ∫ \frac{t^{8} + t^{7} − t^{7} − t^{6} + t^{6} + t^{5} − t^{5} − t^{4} + t^{4} + t^{3} − t^{3} − t^{2} + t^{2} + t − t − 1 + 1}{t + 1} dt =

x > 0, x = t^{12} > 0, t = \sqrt[12]{x}, dx = 12t^{11} dt, t^{8} = \sqrt[12]{x^{8}} = \sqrt[3]{x^{2}}, t^{6} = \sqrt[12]{x^{6}} = \sqrt{x}, t^{4} = \sqrt[12]{x^{4}} = \sqrt[3]{x}, t^{3} = \sqrt[12]{x^{3}} = \sqrt[4]{x}, t^{2} = \sqrt[12]{x^{2}} = \sqrt[6]{x}

= 12 ∫ [t^{7} − t^{6} + t^{5} − t^{4} + t^{3} − t^{2} + t − 1 + \frac{1}{t + 1}] dt = 12 [\frac{t^{8}}{8} − \frac{t^{7}}{7} + \frac{t^{6}}{6} − \frac{t^{5}}{5} + \frac{t^{4}}{4} − \frac{t^{3}}{3} + \frac{t^{2}}{2} − t + ln |1 + t|] + c =

= \frac{3\sqrt[3]{x^{2}}}{2} − \frac{12 \sqrt[12]{x^{7}}}{7} + 2\sqrt{x} − \frac{12 \sqrt[12]{x^{5}}}{5} + 3\sqrt[3]{x} − 4\sqrt[4]{x} + 6\sqrt[6]{x} − 12 \sqrt[12]{x} + 12 ln |1 + \sqrt[12]{x}| + c, for x∈(0 ; ∞)

∫ \frac{dx}{\sqrt[n]{x^{n} + 1}} = ∫ \frac{\sqrt[n]{t^{n} − 1}}{t} \frac{−t^{n−1} dt}{\sqrt[n]{(t^{n} − 1)^{n+1}}} = − ∫ \frac{t^{n−2} dt}{\sqrt[n]{(t^{n} − 1)^{n}}} = − ∫ \frac{t^{n−2} dt}{t^{n} − 1} = ⋯ [Partial fractions], for x∈(0 ; ∞), n = 2, 3, 4, . . ..

x > 0, t = \sqrt[n]{1 + \frac{1}{x^{n}}} = \frac{\sqrt[n]{x^{n} + 1}}{x} > 1, t^{n} = 1 + \frac{1}{x^{n}} ⟹ t^{n}x^{n} = x^{n} + 1 ⟹ x^{n} = \frac{1}{t^{n} − 1} ⟹ x = \frac{1}{\sqrt[n]{t^{n} − 1}}

dx = −\frac{1}{n} (t^{n} − 1)^{− \frac{1}{n} −1} nt^{n−1} dt = \frac{−t^{n−1} dt}{\sqrt[n]{(t^{n} − 1)^{n+1}}} , x^{n} + 1 = \frac{1}{t^{n} − 1} + 1 = \frac{t^{n}}{t^{n} − 1} ⟹ \sqrt[n]{x^{n} + 1} = t (t^{n} − 1)^{− \frac{1}{n}} = \frac{t}{\sqrt[n]{t^{n} − 1}}

∫ \frac{dx}{\sqrt[n]{1 − x^{n}}} = ∫ \frac{\sqrt[n]{t^{n} + 1}}{t} \frac{−t^{n−1} dt}{\sqrt[n]{(t^{n} + 1)^{n+1}}} = − ∫ \frac{t^{n−2} dt}{\sqrt[n]{(t^{n} + 1)^{n}}} = − ∫ \frac{t^{n−2} dt}{t^{n} + 1} = ⋯ [Partial fractions], for x∈(0 ; 1), n = 2, 3, 4, . . ..

0 < x < 1, t = \sqrt[n]{\frac{1}{x^{n}} − 1} = \frac{\sqrt[n]{1 − x^{n}}}{x} > 0, t^{n} = \frac{1}{x^{n}} − 1 ⟹ t^{n}x^{n} = 1 − x^{n} ⟹ x^{n} = \frac{1}{t^{n} + 1} ⟹ x = \frac{1}{\sqrt[n]{t^{n} + 1}}

dx = −\frac{1}{n} (t^{n} + 1)^{− \frac{1}{n} −1} nt^{n−1} dt = \frac{−t^{n−1} dt}{\sqrt[n]{(t^{n} + 1)^{n+1}}} , 1 − x^{n} = 1 − \frac{1}{t^{n} + 1} = \frac{t^{n}}{t^{n} + 1} ⟹ \sqrt[n]{1 − x^{n}} = t (t^{n} + 1)^{− \frac{1}{n}} = \frac{t}{\sqrt[n]{t^{n} + 1}}

∫ \frac{dx}{\sqrt[n]{x^{n} − 1}} = ∫ \frac{\sqrt[n]{1 − t^{n}}}{t} \frac{t^{n−1} dt}{\sqrt[n]{(1 − t^{n})^{n+1}}} = ∫ \frac{t^{n−2} dt}{\sqrt[n]{(1 − t^{n})^{n}}} = − ∫ \frac{t^{n−2} dt}{t^{n} − 1} = ⋯ [Partial fractions], for x∈(1 ; ∞), n = 2, 3, 4, . . ..

x > 1, t = \sqrt[n]{1 − \frac{1}{x^{n}}} = \frac{\sqrt[n]{x^{n} − 1}}{x} ∈ (0 ; 1) , t^{n} = 1 − \frac{1}{x^{n}} ⟹ t^{n}x^{n} = x^{n} − 1 ⟹ x^{n} = \frac{1}{1 − t^{n}} ⟹ x = \frac{1}{\sqrt[n]{1 − t^{n}}}

dx = −\frac{1}{n} (1 − t^{n})^{− \frac{1}{n} −1} (−nt^{n−1}) dt = \frac{t^{n−1} dt}{\sqrt[n]{(1 − t^{n})^{n+1}}} , x^{n} − 1 = \frac{1}{1 − t^{n}} − 1 = \frac{t^{n}}{1 − t^{n}} ⟹ \sqrt[n]{x^{n} − 1} = t (1 − t^{n})^{− \frac{1}{n}} = \frac{t}{\sqrt[n]{1 − t^{n}}}

∫ \frac{dx}{\sqrt{x^{2} + 1}} = − ∫ \frac{dt}{t^{2} − 1} = −\frac{1}{2} ln |\frac{t − 1}{t + 1}| + c = \frac{1}{2} ln \frac{t + 1}{t − 1} + c = \frac{1}{2} ln (\sqrt{x^{2} + 1} + x)^{2} + c = ln (\sqrt{x^{2} + 1} + x) + c, for x > 0.

x > 0, t = \sqrt{1 + \frac{1}{x^{2}}} = \frac{\sqrt{x^{2} + 1}}{x} > 1 ⟹ x = \frac{1}{\sqrt{t^{2} − 1}}, dx = −\frac{t dt}{\sqrt{(t^{2} − 1)^{3}}} , \frac{1}{x^{2} + 1} = \frac{\sqrt{t^{2} − 1}}{t}, \frac{dx}{\sqrt{x^{2} + 1}} = −\frac{dt}{t^{2} − 1}

t ± 1 = \frac{\sqrt{x^{2} + 1}}{x} ± 1 = \frac{\sqrt{x^{2} + 1} ± x}{x} ⟹ \frac{t + 1}{t − 1} = \frac{\sqrt{x^{2} + 1} + x}{\sqrt{x^{2} + 1} − x} = \frac{(\sqrt{x^{2} + 1} + x)^{2}}{(x^{2} + 1) − x^{2}} = (\sqrt{x^{2} + 1} + x)^{2}

∫ \frac{dx}{\sqrt{x^{2} + 1}} = \begin{matrix}x < 0 ⟹ t = −x > 0 \\ dx = − dt, \sqrt{x^{2} + 1} − x > 0 \end{matrix} = − ∫ \frac{dt}{\sqrt{1 + t^{2}}} = − ln (\sqrt{t^{2} + 1} + t) + c = − ln (\sqrt{x^{2} + 1} − x) + c =

= − ln [(\sqrt{x^{2}+1} − x) \frac{\sqrt{x^{2}+1} + x}{\sqrt{x^{2}+1} + x}]+c = − ln \frac{(x^{2} + 1) − x^{2}}{\sqrt{x^{2}+1} + x} +c = − ln \frac{1}{\sqrt{x^{2}+1} + x} +c = ln (\sqrt{x^{2}+1} + x) + c, for x < 0.

⟹ ∫ \frac{dx}{\sqrt{x^{2} + 1}} = ln (\sqrt{x^{2} + 1} + x) + c, for x∈R − \{0\}.

∫ \frac{dx}{\sqrt[3]{x^{3} + 1}} = ∫ \frac{\sqrt[3]{t^{3} − 1}}{t} \frac{−t^{2} dt}{\sqrt[3]{(t^{3} − 1)^{4}}} = −∫ \frac{t dt}{t^{3} − 1} = \frac{t}{t^{3} − 1} = \frac{t}{(t − 1)(t^{2} + t + 1)} = \frac{A}{t − 1} + \frac{Bt + C}{t^{2} + t + 1} ⇒ A = \frac{1}{3}, B = −\frac{1}{3}, C = \frac{1}{3} =

x > 0, t = \sqrt[3]{1 + \frac{1}{x^{3}}} = \frac{\sqrt[3]{x^{3} + 1}}{x} > 1, t^{3} = 1 + \frac{1}{x^{3}} ⟹ t^{3}x^{3} = x^{3} + 1 ⟹ x^{3} = \frac{1}{t^{3} − 1} ⟹ x = \frac{1}{\sqrt[3]{t^{3} − 1}}

dx = −\frac{1}{3} (t^{3} − 1)^{− \frac{4}{3}} 3t^{2} dt = \frac{−t^{2} dt}{\sqrt[3]{(t^{3} − 1)^{4}}} , x^{3} + 1 = \frac{1}{t^{3} − 1} + 1 = \frac{t^{3}}{t^{3} − 1} ⟹ \sqrt[3]{x^{3} + 1} = t (t^{3} − 1)^{− \frac{1}{3}} = \frac{t}{\sqrt[3]{t^{3} − 1}}

\frac{1}{\sqrt[3]{x^{3} + 1}} = \frac{\sqrt[3]{t^{3} − 1}}{t}, t − 1 = \frac{\sqrt[3]{x^{3} + 1}}{x} − 1 = \frac{\sqrt[3]{x^{3} + 1} − x}{x}, t^{3} − 1 = \frac{1}{x^{3}} , 2t + 1 = \frac{2\sqrt[3]{x^{3} + 1}}{x} + 1 = \frac{2\sqrt[3]{x^{3} + 1} + x}{x}

= − ∫ [\frac{\frac{1}{3}}{t − 1} + \frac{−\frac{t}{3} + \frac{1}{3}}{t^{2} + t + 1}] dx = −\frac{1}{3} ∫ \frac{dt}{t − 1} + \frac{1}{6} ∫ \frac{2t − 2}{t^{2} + t + 1} dt = −\frac{1}{3} ∫ \frac{dt}{t − 1} + \frac{1}{6} ∫ \frac{2t + 1}{t^{2} + t + 1} dt − \frac{1}{6} ∫ \frac{3 dt}{t^{2} + t + 1} =

= −\frac{1}{3} ln |t − 1| + \frac{1}{6} ln |t^{2} + t + 1| − \frac{1}{2} ∫ \frac{dt}{t^{2} + t + 1} =_{15}^{page} −\frac{1}{3} ln (t − 1) + \frac{1}{6} ln (t^{2} + t + 1) − \frac{1}{2} \frac{1}{\sqrt{1 − \frac{1}{4}}} arctg \frac{t + \frac{1}{2}}{\sqrt{1 − \frac{1}{4}}} + c =

= −\frac{1}{3} ln (t − 1) − \frac{1}{6} ln (t − 1) + \frac{1}{6} ln (t − 1) + \frac{1}{6} ln (t^{2} + t + 1) − \frac{1}{\sqrt{3}} arctg \frac{2 (t + \frac{1}{2})}{\sqrt{3}} + c =

= −\frac{3}{6} ln (t − 1) + \frac{1}{6} ln [(t − 1)(t^{2} + t + 1)] − \frac{\sqrt{3}}{3} arctg \frac{2t + 1}{\sqrt{3}} + c = −\frac{1}{2} ln (t − 1) + \frac{1}{6} ln (t^{3} − 1) − \frac{\sqrt{3}}{3} arctg \frac{2t + 1}{\sqrt{3}} + c =

= −\frac{1}{2} ln \frac{\sqrt[3]{x^{3} + 1} − x}{x} + \frac{1}{6} ln \frac{1}{x^{3}} − \frac{\sqrt{3}}{3} arctg \frac{2\sqrt[3]{x^{3} + 1} + x}{x\sqrt{3}} +c = −\frac{1}{2} ln \frac{\sqrt[3]{x^{3} + 1} − x}{x} + \frac{1}{6} ln x^{−3}− \frac{\sqrt{3}}{3} arctg \frac{2\sqrt[3]{x^{3} + 1} + x}{x\sqrt{3}} +c =

= −\frac{1}{2} ln (\sqrt[3]{x^{3} + 1} − x) + \frac{1}{2} ln x + \frac{−3}{6} ln x − \frac{\sqrt{3}}{3} arctg \frac{2\sqrt[3]{x^{3} + 1} + x}{x\sqrt{3}} + c =

= −\frac{1}{2} ln (\sqrt[3]{x^{3} + 1} − x) − \frac{\sqrt{3}}{3} arctg \frac{2\sqrt[3]{x^{3} + 1} + x}{x\sqrt{3}} + c, for x∈(0 ; ∞).

∫ \frac{dx}{\sqrt[3]{x^{3} + 1}} = \begin{matrix}−1 < x < 0, u = −x ∈ (0 ; 1) \\ dx = − du, x^{3} + 1 = 1 − u^{3} \end{matrix} = − ∫ \frac{du}{\sqrt[3]{1 − u^{3}}} =_{70}^{page} −\frac{1}{2} ln (\sqrt[3]{1 − u^{3}} + u) + \frac{\sqrt{3}}{3} arctg \frac{2\sqrt[3]{1 − u^{3}} − u}{u\sqrt{3}} + c =

= −\frac{1}{2} ln (\sqrt[3]{x^{3} + 1} − x) + \frac{\sqrt{3}}{3} arctg \frac{2\sqrt[3]{x^{3} + 1} + x}{−x\sqrt{3}} + c = −\frac{1}{2} ln (\sqrt[3]{x^{3} + 1} − x) − \frac{\sqrt{3}}{3} arctg \frac{2\sqrt[3]{x^{3} + 1} + x}{x\sqrt{3}} + c,

for x∈(−1 ; 0).

⟹ ∫ \frac{dx}{\sqrt[3]{x^{3} + 1}} = −\frac{1}{2} ln (\sqrt[3]{x^{3} + 1} − x) − \frac{\sqrt{3}}{3} arctg \frac{2\sqrt[3]{x^{3} + 1} + x}{x\sqrt{3}} + c, for x∈(−1 ; ∞) − \{0\}.

∫ \frac{dx}{\sqrt[4]{x^{4} + 1}} = − ∫ \frac{t^{2} dt}{t^{4} − 1} = \frac{t^{2}}{t^{4} − 1} = \frac{t^{2}}{(t − 1)(t + 1)(t^{2} + 1)} = \frac{A}{t − 1} + \frac{B}{t + 1} + \frac{Ct + D}{t^{2} + 1} ⟹ A = \frac{1}{4}, B = −\frac{1}{4}, C = 0, D = \frac{1}{2} =

x > 0, t = \sqrt[4]{1 + \frac{1}{x^{4}}} = \frac{\sqrt[4]{x^{4} + 1}}{x} > 1, t^{4} = 1 + \frac{1}{x^{4}} ⟹ t^{4}x^{4} = x^{4} + 1 ⟹ x^{4} = \frac{1}{t^{4} − 1} ⟹ x = \frac{1}{\sqrt[4]{t^{4} − 1}}

dx = −\frac{1}{4} (t^{4} − 1)^{− \frac{5}{4}} 4t^{3} dt = \frac{−t^{3} dt}{\sqrt[4]{(t^{4} − 1)^{5}}} , x^{4} + 1 = \frac{1}{t^{4} − 1} + 1 = \frac{t^{4}}{t^{4} − 1} ⟹ \sqrt[4]{x^{4} + 1} = t (t^{4} − 1)^{− \frac{1}{4}} = \frac{t}{\sqrt[4]{t^{4} − 1}}

\frac{1}{\sqrt[4]{x^{4} + 1}} = \frac{\sqrt[4]{t^{4} − 1}}{t}, |t ± 1| = t ± 1 = \frac{\sqrt[4]{x^{4} + 1}}{x} ± 1 = \frac{\sqrt[4]{x^{4} + 1} ± x}{x}, \frac{t + 1}{t − 1} = \frac{\sqrt[4]{x^{4} + 1} + x}{\sqrt[4]{x^{4} + 1} − x}

= − ∫ [\frac{\frac{1}{4}}{t − 1} + \frac{−\frac{1}{4}}{t + 1} + \frac{\frac{1}{2}}{t^{2} + 1}] dt = −\frac{1}{4} ln |t − 1| + \frac{1}{4} ln |t + 1| − \frac{1}{2} arctg t + c = \frac{1}{4} ln \frac{t + 1}{t − 1} − \frac{1}{2} arctg t + c =

= \frac{1}{4} ln \frac{\sqrt[4]{x^{4} + 1} + x}{\sqrt[4]{x^{4} + 1} − x} − \frac{1}{2} arctg \frac{\sqrt[4]{x^{4} + 1}}{x} + c, for x∈(0 ; ∞).

∫ \frac{dx}{\sqrt[4]{x^{4} + 1}} = x < 0, x = −u > 0, dx = − du = − ∫ \frac{du}{\sqrt[4]{u^{4} + 1}} = −\frac{1}{4} ln \frac{\sqrt[4]{u^{4} + 1} + u}{\sqrt[4]{u^{4} + 1} − u} + \frac{1}{2} arctg \frac{\sqrt[4]{u^{4} + 1}}{u} + c =

= −\frac{1}{4} ln \frac{\sqrt[4]{x^{4} + 1} − x}{\sqrt[4]{x^{4} + 1} + x} + \frac{1}{2} arctg \frac{\sqrt[4]{x^{4} + 1}}{−x} + c = \frac{1}{4} ln \frac{\sqrt[4]{x^{4} + 1} + x}{\sqrt[4]{x^{4} + 1} − x} − \frac{1}{2} arctg \frac{\sqrt[4]{x^{4} + 1}}{x} + c, for x∈(−∞; 0).

⟹ ∫ \frac{dx}{\sqrt[4]{x^{4} + 1}} = \frac{1}{4} ln \frac{\sqrt[4]{x^{4} + 1} + x}{\sqrt[4]{x^{4} + 1} − x} − \frac{1}{2} arctg \frac{\sqrt[4]{x^{4} + 1}}{x} + c, for x∈R − \{0\}.

∫ \frac{dx}{\sqrt{x^{2} − 1}} = ∫ \frac{\sqrt{1 − t^{2}}}{t} \frac{t dt}{\sqrt{(1 − t^{2})^{3}}} = − ∫ \frac{dt}{t^{2} − 1} = −\frac{1}{2} ln |\frac{t − 1}{t + 1}| + c = \frac{1}{2} ln |\frac{t + 1}{t − 1}| + c =

x > 1, t = \sqrt{1 − \frac{1}{x^{2}}} = \frac{\sqrt{x^{2} − 1}}{x} ∈ (0 ; 1) ⟹ x = \frac{1}{\sqrt{1 − t^{2}}} , dx = \frac{t dt}{\sqrt{(1 − t^{2})^{3}}} , x^{2} − 1 = \frac{t^{2}}{1 − t^{2}} , \frac{1}{\sqrt{x^{2} − 1}} = \frac{\sqrt{1 − t^{2}}}{t}

t ± 1 = \frac{\sqrt{x^{2} − 1}}{x} ± 1 = \frac{\sqrt{x^{2} − 1} ± x}{x} ⟹ \frac{t + 1}{t − 1} = \frac{\sqrt{x^{2} − 1} + x}{\sqrt{x^{2} − 1} − x} = \frac{(\sqrt{x^{2} − 1} + x)^{2}}{(x^{2} − 1) − x^{2}} = − (\sqrt{x^{2} − 1} + x)^{2}

= \frac{1}{2} ln |(\sqrt{x^{2} − 1} + x)^{2}| + c = \frac{2}{2} ln |\sqrt{x^{2} − 1} + x| + c = ln |\sqrt{x^{2} − 1} + x| + c, for x > 1.

∫ \frac{dx}{\sqrt{x^{2} − 1}} = x < −1, u = −x > 1, dx = − dt = − ∫ \frac{du}{\sqrt{u^{2} − 1}} = − ln |\sqrt{u^{2} − 1} + u| + c = − ln |\sqrt{x^{2} − 1} − x| + c =

= − ln |(\sqrt{x^{2} − 1} − x) \frac{\sqrt{x^{2} − 1} + x}{\sqrt{x^{2} − 1} + x}| + c = − ln |\frac{(x^{2} − 1) − x^{2}}{\sqrt{x^{2} − 1} + x} | + c = − ln |\frac{1}{\sqrt{x^{2} − 1} + x}| + c = ln |\sqrt{x^{2} − 1} + x| + c,

for x < −1.

⟹ ∫ \frac{dx}{\sqrt{x^{2} − 1}} = ln |\sqrt{x^{2} − 1} + x| + c, for x∈R − ⟨−1 ; 1⟩ = (−∞; −1) ∪ (1 ; ∞).

∫ \frac{dx}{\sqrt[3]{x^{3} − 1}} = ∫ \frac{t dt}{1 − t^{3}} = ∫ [\frac{−\frac{1}{3}}{t − 1} + \frac{\frac{t}{3} − \frac{1}{3}}{t^{2} + t + 1}] dt = −\frac{1}{3} ∫ \frac{dt}{t − 1} + \frac{1}{6} ∫ \frac{2t − 2}{t^{2} + t + 1} dt = −\frac{1}{3} ln |t − 1| + \frac{1}{6} ∫ \frac{2t + 1 − 3}{t^{2} + t + 1} dt =

x > 1, t = \sqrt[3]{1 − \frac{1}{x^{3}}} = \frac{\sqrt[3]{x^{3} − 1}}{x} ∈ (0 ; 1) , t^{3} = 1 − \frac{1}{x^{3}} ⟹ 1 − t^{3} = \frac{1}{x^{3}} , x^{3} = \frac{1}{1 − t^{3}} ⟹ x = \frac{1}{\sqrt[3]{1 − t^{3}}} , dx = \frac{t^{2} dt}{\sqrt[3]{(1 − t^{3})^{4}}}

x^{3} − 1 = \frac{t^{3}}{1 − t^{3}} ⟹ \sqrt[3]{x^{3} − 1} = \frac{t}{\sqrt[3]{1 − t^{3}}} , 1 − t = 1 − \frac{\sqrt[3]{x^{3} − 1}}{x} = \frac{x − \sqrt[3]{x^{3} − 1}}{x}, 2t + 1 = \frac{2\sqrt[3]{x^{3} − 1}}{x} + 1 = \frac{2\sqrt[3]{x^{3} − 1} + x}{x}

\frac{t}{1 − t^{3}} = \frac{−t}{(t − 1)(t^{2} + t + 1)} = \frac{A}{t − 1} + \frac{Bt + C}{t^{2} + t + 1} ⟹ A = −\frac{1}{3}, B = \frac{1}{3}, C = −\frac{1}{3}

= −\frac{1}{3} ln |t − 1| + \frac{1}{6} ∫ \frac{2t + 1}{t^{2} + t + 1} dt − \frac{1}{2} ∫ \frac{dt}{t^{2} + t + 1} =_{15}^{page} −\frac{1}{3} ln |t − 1| + \frac{1}{6} ln |t^{2} + t + 1| − \frac{1}{2} \frac{1}{\sqrt{1 − \frac{1}{4}}} arctg \frac{t + \frac{1}{2}}{\sqrt{1 − \frac{1}{4}}} + c =

= −\frac{1}{3} ln (1 − t) − \frac{1}{6} ln (1 − t) + \frac{1}{6} ln (1 − t) + \frac{1}{6} ln (t^{2} + t + 1) − \frac{1}{\sqrt{3}} arctg \frac{2(t + \frac{1}{2})}{\sqrt{3}} + c =

= −\frac{3}{6} ln (1 − t) + \frac{1}{6} ln [(1 − t)(t^{2} + t + 1)] − \frac{\sqrt{3}}{3} arctg \frac{2t + 1}{\sqrt{3}} + c = −\frac{1}{2} ln (1 − t) + \frac{1}{6} ln (1 − t^{3}) − \frac{\sqrt{3}}{3} arctg \frac{2t + 1}{\sqrt{3}} + c =

= −\frac{1}{2} ln \frac{x−\sqrt[3]{x^{3}−1}}{x} + \frac{1}{6} ln \frac{1}{x^{3}} − \frac{1}{\sqrt{3}} arctg \frac{2\sqrt[3]{x^{3}−1} + x}{x\sqrt{3}} +c = −\frac{1}{2} ln (x−\sqrt[3]{x^{3}−1})+ \frac{1}{2} ln x− \frac{1}{6} ln x^{3}− \frac{1}{\sqrt{3}} arctg \frac{2\sqrt[3]{x^{3}−1} + x}{x\sqrt{3}} +c =

= −\frac{1}{2} ln (x − \sqrt[3]{x^{3} − 1}) − \frac{1}{\sqrt{3}} arctg \frac{2\sqrt[3]{x^{3} − 1} + x}{x\sqrt{3}} + c, for x∈(1 ; ∞).

∫ \frac{dx}{\sqrt[4]{x^{4} − 1}} = ∫ \frac{t^{2} dt}{1 − t^{4}} = ∫ [\frac{−\frac{1}{4}}{t − 1} + \frac{\frac{1}{4}}{t + 1} + \frac{−\frac{1}{2}}{t^{2} + 1}] dt = −\frac{1}{4} ln |t − 1| + \frac{1}{4} ln |t + 1| − \frac{1}{2} arctg t + c =

x > 1, t = \sqrt[4]{1 − \frac{1}{x^{4}}} = \frac{\sqrt[4]{x^{4} − 1}}{x} ∈ (0 ; 1) , t^{4} = 1 − \frac{1}{x^{4}} ⟹ x^{4} = \frac{1}{1 − t^{4}} , x = \frac{1}{\sqrt[4]{1 − t^{4}}} , dx = \frac{t^{3} dt}{\sqrt[4]{(1 − t^{4})^{5}}} ,

x^{4} − 1 = \frac{t^{4}}{1 − t^{4}} , \sqrt[4]{x^{4} − 1} = \frac{t}{\sqrt[4]{1 − t^{4}}} , |1 ± t| = 1 ± t = 1 ± \frac{\sqrt[4]{x^{4} − 1}}{x} = \frac{x ± \sqrt[4]{x^{4} − 1}}{x}, \frac{1 + t}{1 − t} = \frac{x + \sqrt[4]{x^{4} − 1}}{x − \sqrt[4]{x^{4} − 1}}

\frac{t^{2}}{1 − t^{4}} = \frac{−t^{2}}{(t − 1)(t + 1)(t^{2} + 1)} = \frac{A}{t − 1} + \frac{B}{t + 1} + \frac{Ct + D}{t^{2} + 1} ⟹ A = −\frac{1}{4}, B = \frac{1}{4}, C = 0, D = −\frac{1}{2}

= \frac{1}{4} ln |\frac{t + 1}{t − 1}| − \frac{1}{2} arctg t + c = \frac{1}{4} ln \frac{1 + t}{1 − t} − \frac{1}{2} arctg t + c = \frac{1}{4} ln \frac{x + \sqrt[4]{x^{4} − 1}}{x − \sqrt[4]{x^{4} − 1}} − \frac{1}{2} arctg \frac{\sqrt[4]{x^{4} − 1}}{x} + c, for x∈(1 ; ∞).

∫ \frac{dx}{\sqrt[4]{x^{4} − 1}} = x < −1, u = −x > 1, dx = − du = − ∫ \frac{du}{\sqrt[4]{u^{4} − 1}} = −\frac{1}{4} ln \frac{u + \sqrt[4]{u^{4} − 1}}{u − \sqrt[4]{u^{4} − 1}} + \frac{1}{2} arctg \frac{\sqrt[4]{u^{4} − 1}}{u} + c =

= −\frac{1}{4} ln \frac{−x + \sqrt[4]{x^{4} − 1}}{−x − \sqrt[4]{x^{4} − 1}} + \frac{1}{2} arctg \frac{\sqrt[4]{x^{4} − 1}}{−x} + c = −\frac{1}{4} ln \frac{x − \sqrt[4]{x^{4} − 1}}{x + \sqrt[4]{x^{4} − 1}} − \frac{1}{2} arctg \frac{\sqrt[4]{x^{4} − 1}}{x} + c =

= \frac{1}{4} ln \frac{x + \sqrt[4]{x^{4} − 1}}{x − \sqrt[4]{x^{4} − 1}} − \frac{1}{2} arctg \frac{\sqrt[4]{x^{4} − 1}}{x} + c, for x∈(−∞; −1).

⟹ ∫ \frac{dx}{\sqrt[4]{x^{4} − 1}} = \frac{1}{4} ln \frac{x + \sqrt[4]{x^{4} − 1}}{x − \sqrt[4]{x^{4} − 1}} − \frac{1}{2} arctg \frac{\sqrt[4]{x^{4} − 1}}{x} + c, for x∈(−∞; −1) ∪ (1 ; ∞).

∫ \frac{dx}{\sqrt{1 − x^{2}}} = ∫ \frac{\sqrt{t^{2} + 1}}{t} \frac{−t dt}{\sqrt{(t^{2} + 1)^{3}}} = − ∫ \frac{dt}{t^{2} + 1} = − arctg t + c_{1} = − arctg \frac{\sqrt{1 − x^{2}}}{x} + c_{1}, for x∈(0 ; 1).

0 < x < 1, t = \sqrt{\frac{1}{x^{2}} − 1} = \frac{\sqrt{1 − x^{2}}}{x} > 0, t^{2} = \frac{1}{x^{2}} − 1 ⟹ t^{2}x^{2} = 1 − x^{2} ⟹ x^{2} = \frac{1}{t^{2} + 1} ⟹ x = \frac{1}{\sqrt{t^{2} + 1}}

dx = −\frac{1}{2} (t^{2} + 1)^{− \frac{3}{2}} 2t dt = \frac{−t dt}{\sqrt{(t^{2} + 1)^{3}}} , 1 − x^{2} = 1 − \frac{1}{t^{2} + 1} = \frac{t^{2}}{t^{2} + 1}, \sqrt{1 − x^{2}} = \frac{t}{\sqrt{t^{2} + 1}}, \frac{1}{x^{2} + 1} = \frac{\sqrt{t^{2} + 1}}{t}

∫ \frac{dx}{\sqrt{1 − x^{2}}} = \begin{matrix}−1 < x < 0, u = −x > 0 \\ dx = − du \end{matrix} = − ∫ \frac{du}{\sqrt{1 − u^{2}}} = arctg \frac{\sqrt{1 − u^{2}}}{u} + c_{1} = arctg \frac{\sqrt{1 − x^{2}}}{−x} + c_{1} = − arctg \frac{\sqrt{1 − x^{2}}}{x} + c_{1},

for x∈(−1 ; 0).

⟹ ∫ \frac{dx}{\sqrt{1 − x^{2}}} = − arctg \frac{\sqrt{1 − x^{2}}}{x} + c_{1}, for x∈(−1 ; 1) − \{0\},

∫ \frac{dx}{\sqrt{1 − x^{2}}} = 2-nd Euler: − 1 < x < 1, \sqrt{1 − x^{2}} = xt + 1 ⇒ 1 − x^{2} = x^{2}t^{2} + 2tx + 1 ⇒ −2tx = x^{2} + x^{2}t^{2} ⟹^{x≠0} x = \frac{−2t}{t^{2} + 1} t = \frac{\sqrt{1 − x^{2}} − 1}{x}, \sqrt{1 − x^{2}} = \frac{−2t^{2}}{t^{2} + 1} + 1 = \frac{1 − t^{2}}{t^{2} + 1}, dx = \frac{−2(t^{2} + 1) + 2t · 2t}{(t^{2} + 1)^{2}} dt = \frac{2(t^{2} − 1)}{(t^{2} + 1)^{2}} dt =

= ∫ \frac{t^{2} + 1}{1 − t^{2}} \frac{2(t^{2} − 1)}{(t^{2} + 1)^{2}} dt = −2 ∫ \frac{dt}{t^{2} + 1} = −2 arctg t + c_{2} = −2 arctg \frac{\sqrt{1 − x^{2}} − 1}{x} + c_{2}, for x∈(−1 ; 1) − \{0\}.

∫ \frac{dx}{\sqrt{1 − x^{2}}} = −1 < x < 1, x = sin t, t = arcsin x ∈ (−π/2 ; π/2) dx = cos t dt, \sqrt{1−x^{2}} = \sqrt{1−sin^{2} t} = \sqrt{cos^{2} t} = |cos t| = cos t =∫ \frac{cos t dt}{cos t} =∫ dt = t+c_{3} = arcsin x + c_{3},

for x∈(−1 ; 1).

∫ \frac{dx}{\sqrt{1 − x^{2}}} = −1 < x < 1, x = cos t, t = arccos x ∈ (0 ; π) , dx = − sin t dt \sqrt{1−x^{2}} = \sqrt{1−cos^{2} t} = \sqrt{sin^{2} t} = |sin t| = sin t, t ∈ (0 ; π) = − ∫ \frac{sin t dt}{sin t} = − ∫ dt = −t+c_{4} = − arccos x + c_{4},

for x∈(−1 ; 1).

⟹ ∫ \frac{dx}{\sqrt{1 − x^{2}}} = − arctg \frac{\sqrt{1 − x^{2}}}{x} + c_{1} = −2 arctg \frac{\sqrt{1 − x^{2}} − 1}{x} + c_{2}, for x∈(−1 ; 1) − \{0\},

= arcsin x + c_{3} = − arccos x + c_{4}, for x∈(−1 ; 1).

∫ \frac{dx}{\sqrt[3]{1 − x^{3}}} = − ∫ \frac{t dt}{t^{3} + 1} = ∫ [\frac{\frac{1}{3}}{t + 1} + \frac{−\frac{t}{3} − \frac{1}{3}}{t^{2} − t + 1}] dt = \frac{1}{3} ∫ \frac{dt}{t + 1} − \frac{1}{6} ∫ \frac{2t + 2}{t^{2} − t + 1} dt = \frac{1}{3} ln |t + 1| − \frac{1}{6} ∫ \frac{2t − 1 + 3}{t^{2} − t + 1} dt =

0 < x < 1, t = \sqrt[3]{\frac{1}{x^{3}} − 1} = \frac{\sqrt[3]{1 − x^{3}}}{x} > 0, t^{3} = \frac{1}{x^{3}} − 1 ⟹ t^{3} + 1 = \frac{1}{x^{3}} , x^{3} = \frac{1}{t^{3} + 1}, x = \frac{1}{\sqrt[3]{t^{3} + 1}}, dx = \frac{−t^{2} dt}{\sqrt[3]{(t^{3} + 1)^{4}}}

1 − x^{3} = \frac{t^{3}}{t^{3} + 1}, \sqrt[3]{1 − x^{3}} = \frac{t}{\sqrt[3]{t^{3} + 1}}, t + 1 = \frac{\sqrt[3]{1 − x^{3}}}{x} + 1 = \frac{\sqrt[3]{1 − x^{3}} + x}{x}, 2t − 1 = \frac{2\sqrt[3]{1 − x^{3}}}{x} − 1 = \frac{2\sqrt[3]{1 − x^{3}} − x}{x}

−\frac{t}{t^{3} + 1} = \frac{−t}{(t + 1)(t^{2} − t + 1)} = \frac{A}{t + 1} + \frac{Bt + C}{t^{2} − t + 1} ⟹ A = \frac{1}{3}, B = −\frac{1}{3}, C = −\frac{1}{3}

= \frac{1}{3} ln (t + 1) − \frac{1}{6} ∫ \frac{2t − 1}{t^{2} − t + 1} dt − \frac{1}{2} ∫ \frac{dt}{t^{2} − t + 1} =_{15}^{page} \frac{1}{3} ln (t + 1) − \frac{1}{6} ln |t^{2} − t + 1| − \frac{1}{2} \frac{1}{\sqrt{1 − \frac{1}{4}}} arctg \frac{t − \frac{1}{2}}{\sqrt{1 − \frac{1}{4}}} + c =

= \frac{1}{3} ln (t + 1) − \frac{1}{6} ln (t^{2} − t + 1) − \frac{1}{\sqrt{3}} arctg \frac{2(t − \frac{1}{2})}{\sqrt{3}} + c = \frac{1}{3} ln (t + 1) − \frac{1}{6} ln \frac{t^{3} + 1}{t + 1} − \frac{\sqrt{3}}{3} arctg \frac{2t − 1}{\sqrt{3}} + c =

= \frac{1}{3} ln (t + 1) − \frac{1}{6} ln (t^{3} + 1) + \frac{1}{6} ln (t+1) − \frac{\sqrt{3}}{3} arctg \frac{2t − 1}{\sqrt{3}} + c = \frac{3}{6} ln (t + 1) − \frac{1}{6} ln (t^{3} + 1) − \frac{\sqrt{3}}{3} arctg \frac{2t − 1}{\sqrt{3}} + c =

= \frac{1}{2} ln \frac{\sqrt[3]{1 − x^{3}} + x}{x} − \frac{1}{6} ln \frac{1}{x^{3}} − \frac{\sqrt{3}}{3} arctg \frac{2\sqrt[3]{1 − x^{3}} − x}{x\sqrt{3}} +c = \frac{1}{2} ln (\sqrt[3]{1 − x^{3}} + x)− \frac{1}{2} ln x+ \frac{1}{6} ln x^{3}− \frac{\sqrt{3}}{3} arctg \frac{2\sqrt[3]{1 − x^{3}} − x}{x\sqrt{3}} +c =

= \frac{1}{2} ln (\sqrt[3]{1 − x^{3}} + x) − \frac{\sqrt{3}}{3} arctg \frac{2\sqrt[3]{1 − x^{3}} − x}{x\sqrt{3}} + c, for x∈(0 ; 1).

∫ \frac{dx}{\sqrt[3]{1 − x^{3}}} = \begin{matrix}x < 0, u = −x > 0, dx = − du \\ 1 − x^{3} = 1 − (−u)^{3} = u^{3} + 1 \end{matrix} = − ∫ \frac{du}{\sqrt[3]{u^{3} + 1}} =_{66}^{page} \frac{1}{2} ln (\sqrt[3]{u^{3} + 1} − u) + \frac{1}{\sqrt{3}} arctg \frac{2\sqrt[3]{u^{3} + 1} + u}{u\sqrt{3}} + c =

= \frac{1}{2} ln (\sqrt[3]{1 − x^{3}} + x) + \frac{1}{\sqrt{3}} arctg \frac{2\sqrt[3]{1 − x^{3}} − x}{−x\sqrt{3}} + c = \frac{1}{2} ln (\sqrt[3]{1 − x^{3}} + x) − \frac{1}{\sqrt{3}} arctg \frac{2\sqrt[3]{1 − x^{3}} − x}{x\sqrt{3}} + c, for x∈(−∞; 0).

⟹ ∫ \frac{dx}{\sqrt[3]{1 − x^{3}}} = \frac{1}{2} ln (\sqrt[3]{1 − x^{3}} + x) − \frac{1}{\sqrt{3}} arctg \frac{2\sqrt[3]{1 − x^{3}} − x}{x\sqrt{3}} + c, for x∈(−∞; 1) − \{0\}.

∫ \frac{dx}{\sqrt[4]{1 − x^{4}}} = ∫ \frac{−t^{2} dt}{t^{4} + 1} = ∫ [\frac{\frac{\sqrt{2}}{4}t}{t^{2} + \sqrt{2}t + 1} + \frac{−\frac{\sqrt{2}}{4}t}{t^{2} − \sqrt{2}t + 1}] dt = \frac{\sqrt{2}}{8} ∫ [\frac{2t}{t^{2} + \sqrt{2}t + 1} − \frac{2t}{t^{2} − \sqrt{2}t + 1}] dt =

0 < x < 1, t = \sqrt[4]{\frac{1}{x^{4}} − 1} = \frac{\sqrt[4]{1 − x^{4}}}{x} > 0, t^{4} = \frac{1}{x^{4}} − 1 ⟹ x^{4} = \frac{1}{t^{4} + 1}, x = \frac{1}{\sqrt[4]{t^{4} + 1}}, dx = \frac{−t^{3} dt}{\sqrt[4]{(t^{4} + 1)^{5}}} , \sqrt[4]{1 − x^{4}} = \frac{t}{\sqrt[4]{t^{4} + 1}}

|t^{2} ± \sqrt{2}t + 1| = t^{2} ± \sqrt{2}t + 1 = \frac{\sqrt{1 − x^{4}}}{x^{2}} ± \sqrt{2} \frac{\sqrt[4]{1 − x^{4}}}{x} + 1 = \frac{\sqrt{1 − x^{4}} ± \sqrt{2}x\sqrt[4]{1 − x^{4}} + x^{2}}{x^{2}} = \frac{\sqrt{1 − x^{4}} ± x\sqrt[4]{4(1 − x^{4})} + x^{2}}{x^{2}}

\frac{t^{2} + \sqrt{2}t + 1}{t^{2} − \sqrt{2}t + 1} = \frac{\sqrt{1 − x^{4}} + x\sqrt[4]{4(1 − x^{4})} + x^{2}}{\sqrt{1 − x^{4}} − x\sqrt[4]{4(1 − x^{4})} + x^{2}} , \sqrt{2}t ± 1 = \sqrt{2} \frac{\sqrt[4]{1 − x^{4}}}{x} ± 1 = \frac{\sqrt{2}\sqrt[4]{1 − x^{4}} ± x}{x} = \frac{\sqrt[4]{4(1 − x^{4})} ± x}{x}

\frac{−t^{2}}{t^{4} + 1} = \frac{−t^{2}}{(t^{2} + \sqrt{2}t + 1)(t^{2} − \sqrt{2}t + 1)} = \frac{At + B}{t^{2} + \sqrt{2}t + 1} + \frac{Ct + D}{t^{2} − \sqrt{2}t + 1} ⟹ A = \frac{\sqrt{2}}{4}, B = 0, C = −\frac{\sqrt{2}}{4}, D = 0

= \frac{\sqrt{2}}{8}∫ [\frac{2t+\sqrt{2}−\sqrt{2}}{t^{2}+\sqrt{2}t+1} − \frac{2t−\sqrt{2}+\sqrt{2}}{t^{2}−\sqrt{2}t+1}] dt = \frac{\sqrt{2}}{8}∫ [\frac{2t + \sqrt{2}}{t^{2}+\sqrt{2}t+1} − \frac{2t − \sqrt{2}}{t^{2}−\sqrt{2}t+1}] dt + \frac{2}{8} ∫ [\frac{−1}{t^{2}+\sqrt{2}t+1} − \frac{1}{t^{2}−\sqrt{2}t+1}] dt =

= \frac{\sqrt{2}}{8} [ln |t^{2} + \sqrt{2}t + 1| − ln |t^{2} − \sqrt{2}t + 1|] − \frac{1}{4} ∫ \frac{dt}{t^{2} + \sqrt{2}t + 1} − \frac{1}{4} ∫ \frac{dt}{t^{2} − \sqrt{2}t + 1} =_{15}^{page}

= \frac{\sqrt{2}}{8} ln |\frac{t^{2} + \sqrt{2}t + 1}{t^{2} − \sqrt{2}t + 1}| − \frac{1}{4} \frac{1}{\sqrt{1 − \frac{2}{4}}} arctg \frac{t + \frac{\sqrt{2}}{2}}{\sqrt{1 − \frac{2}{4}}} − \frac{1}{4} \frac{1}{\sqrt{1 − \frac{2}{4}}} arctg \frac{t − \frac{\sqrt{2}}{2}}{\sqrt{1 − \frac{2}{4}}} + c =

= \frac{\sqrt{2}}{8} ln \frac{t^{2}+\sqrt{2}t+1}{t^{2}−\sqrt{2}t+1} − \frac{1}{2\sqrt{2}} [arctg \frac{t+ \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} + arctg \frac{t− \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} ] + c = \frac{\sqrt{2}}{8} ln \frac{t^{2}+\sqrt{2}t+1}{t^{2}−\sqrt{2}t+1} − \frac{\sqrt{2}}{4} [arctg (\sqrt{2}t+1) + arctg (\sqrt{2}t−1)] + c =

= \frac{\sqrt{2}}{8} ln \frac{\sqrt{1 − x^{4}} + x\sqrt[4]{4(1 − x^{4})} + x^{2}}{\sqrt{1 − x^{4}} − x\sqrt[4]{4(1 − x^{4})} + x^{2}} − \frac{\sqrt{2}}{4} arctg \frac{\sqrt[4]{4(1 − x^{4})} + x}{x} − \frac{\sqrt{2}}{4} arctg \frac{\sqrt[4]{4(1 − x^{4})} − x}{x} + c, for x∈(0 ; 1).

∫ \frac{dx}{\sqrt[4]{1 − x^{4}}} = x < −1, u = −x > 1, dx = − du = − ∫ \frac{du}{\sqrt[4]{1 − u^{4}}} =

= −\frac{\sqrt{2}}{8} ln \frac{\sqrt{1 − u^{4}} + u\sqrt[4]{4(1 − u^{4})} + u^{2}}{\sqrt{1 − u^{4}} − u\sqrt[4]{4(1 − u^{4})} + u^{2}} + \frac{\sqrt{2}}{4} arctg \frac{\sqrt[4]{4(1 − u^{4})} + u}{u} + \frac{\sqrt{2}}{4} arctg \frac{\sqrt[4]{4(1 − u^{4})} − u}{u} + c =

= −\frac{\sqrt{2}}{8} ln \frac{\sqrt{1 − x^{4}} − x\sqrt[4]{4(1 − x^{4})} + x^{2}}{\sqrt{1 − x^{4}} + x\sqrt[4]{4(1 − x^{4})} + x^{2}} + \frac{\sqrt{2}}{4} arctg \frac{\sqrt[4]{4(1 − x^{4})} − x}{−x} + \frac{\sqrt{2}}{4} arctg \frac{\sqrt[4]{4(1 − x^{4})} + x}{−x} + c =

= \frac{\sqrt{2}}{8} ln \frac{\sqrt{1 − x^{4}} + x\sqrt[4]{4(1 − x^{4})} + x^{2}}{\sqrt{1 − x^{4}} − x\sqrt[4]{4(1 − x^{4})} + x^{2}} − \frac{\sqrt{2}}{4} arctg \frac{\sqrt[4]{4(1 − x^{4})} − x}{x} − \frac{\sqrt{2}}{4} arctg \frac{\sqrt[4]{4(1 − x^{4})} + x}{x} + c, for x∈(−1 ; 0).

⟹ ∫ \frac{dx}{\sqrt[4]{1 − x^{4}}} = \frac{\sqrt{2}}{8} ln \frac{\sqrt{1 − x^{4}} + x\sqrt[4]{4(1 − x^{4})} + x^{2}}{\sqrt{1 − x^{4}} − x\sqrt[4]{(1 − x^{4})} + x^{2}} − \frac{\sqrt{2}}{4} arctg \frac{\sqrt[4]{4(1 − x^{4})} + x}{x} − \frac{\sqrt{2}}{4} arctg \frac{\sqrt[4]{4(1 − x^{4})} − x}{x} + c, for x∈(−1 ; 1).

∫ sin nx dx = nx = t, dx = \frac{dt}{n} = \frac{1}{n} ∫ sin t dt = −\frac{cos t}{n} + c = −\frac{cos nx}{n} + c, for n∈N, x∈R.

∫ x sin nx dx = \begin{matrix}u = x ⇒ & u′ = 1 \\ v′ = sin nx ⇒ & v = −\frac{cos nx}{n} \end{matrix} = −\frac{x cos nx}{n} + ∫ \frac{cos nx}{n} dx = −\frac{x cos nx}{n} + \frac{1}{n} \frac{sin nx}{n} + c = \frac{sin nx}{n^{2}} − \frac{x cos nx}{n} + c,

for x∈R, n∈N.

∫ x sin nx dx = [A+Bx] sin nx + [C+Dx] cos nx + c = [\frac{1}{n^{2}} +0] sin nx + [0− \frac{x}{n}] cos nx + c = \frac{sin nx}{n^{2}} − \frac{x cos nx}{n} + c, for x∈R, n∈N.

⟹^{Derivation}

[∫ x sin nx dx]′ = [A + Bx]′ sin nx + [A + Bx] sin′ nx + [C + Dx]′ cos nx + [C + Dx] cos′ nx
x sin nx = B sin nx + [A + Bx]n cos nx + D cos nx − [C + Dx]n sin nx =

= [B − Cn − Dnx] sin nx + [An + D + Bnx] cos nx

⟹^{Equations}

0 = B − Cn, x = −Dnx, 0 = D + An, 0 = Bnx ⟹ A = \frac{1}{n^{2}} , B = 0, C = 0, D = −\frac{1}{n}

∫ x^{2} sin nx dx = \begin{matrix}u = x^{2}  ⇒ & u′ = 2x \\ v′ = sin nx ⇒ & v = −\frac{cos nx}{n} \end{matrix} = −\frac{x^{2} cos nx}{n} + \frac{2}{n} ∫ x cos nx dx = \begin{matrix}u = x  ⇒ & u′ = 1 \\ v′ = cos nx ⇒ & v = \frac{sin nx}{n} \end{matrix} =

= −\frac{x^{2} cos nx}{n} + \frac{2}{n} [\frac{x sin nx}{n} − \frac{1}{n} ∫ sin nx dx] = −\frac{x^{2} cos nx}{n} + \frac{2}{n} [\frac{x sin nx}{n} − \frac{1}{n} \frac{− cos nx}{n}] = \frac{2x sin nx}{n^{2}} + \frac{2 cos nx}{n^{3}} − \frac{x^{2} cos nx}{n} + c,

for x∈R, n∈N.

∫ x^{2} sin nx dx = [A+Bx+Cx^{2}] sin nx + [D+Ex+F x^{2}] cos nx + c = \frac{2x}{n^{2}} sin nx + [\frac{2}{n^{3}} − \frac{x^{2}}{n} ] cos nx + c, for x∈R, n∈N.

⟹^{Derivation}

[∫ x^{2} sin nx dx]′ = [B + 2Cx] sin nx + [A + Bx + Cx^{2}]n cos nx + [E + 2F x] cos nx − [D + Ex + F x^{2}]n sin nx
x^{2} sin nx = [B − Dn + (2C − En)x − F nx^{2}] sin nx + [An + E + (Bn + 2F)x + Cnx^{2}] cos nx

⟹^{Equations} 0 = B−Dn, 0 = (2C−En)x, x^{2} = −F nx^{2}, 0 = E+An, 0 = (2F +Bn)x, 0 = Cnx^{2}

⟹ A = 0, B = \frac{2}{n^{2}} , C = 0, D = \frac{2}{n^{3}} , E = 0, F = −\frac{1}{n}

∫ x^{3} sin nx dx = [A+Bx+Cx^{2}+Dx^{3}] sin nx + [E+F x+Gx^{2}+Hx^{3}] cos nx + c = [−\frac{6}{n^{4}} + \frac{3x^{2}}{n^{2}} ] sin nx + [\frac{6x}{n^{3}} − \frac{x^{3}}{n} ] cos nx + c,

for x∈R, n∈N.

⟹^{Derivation}

[∫ x^{3} sin nx dx]′ = [B + 2Cx + 3Dx^{2}] sin nx + [A + Bx + Cx^{2} + Dx^{3}]n cos nx+

+[F + 2Gx + 3Hx^{2}] cos nx − [E + F x + Gx^{2} + Hx^{3}]n sin nx

x^{3} sin nx = [B − En + (2C − F n)x − (3D − Gn)x^{2} − Hnx^{3}] sin nx + [An + F + (Bn + 2G)x + (Cn + 3H)x^{2} + Dnx^{3}] cos nx

⟹^{Equations} 0=B−En, 0=(2C−F n)x, 0=(3D−Gn)x^{2}, x^{3} =−Hnx^{3}, 0=F +An, 0=(2G+Bn)x, 0=(3H +Cn)x^{2}, 0=Dnx^{3}

⟹ A = −\frac{6}{n^{4}} , B = 0, C = \frac{3}{n^{2}} , D = 0, E = 0, F = \frac{6}{n^{3}} , G = 0, H = −\frac{1}{n}

∫ cos nx dx = nx = t, dx = \frac{dt}{n} = \frac{1}{n} ∫ cos t dt = \frac{sin t}{n} + c = \frac{sin nx}{n} + c, for n∈N, x∈R.

∫ x cos nx dx = \begin{matrix}u = x ⇒ & u′ = 1 \\ v′ = cos nx ⇒ & v = \frac{sin nx}{n} \end{matrix} = \frac{x sin nx}{n} − ∫ \frac{sin nx}{n} dx = \frac{x sin nx}{n} − \frac{1}{n} \frac{− cos nx}{n} + c = \frac{cos nx}{n^{2}} + \frac{x sin nx}{n} + c,

for x∈R, n∈N.

∫ x cos nx dx = [A+Bx] cos nx + [C+Dx] sin nx + c = [\frac{1}{n^{2}} +0] cos nx + [0+ \frac{x}{n}] sin nx + c = \frac{cos nx}{n^{2}} + \frac{x sin nx}{n} + c, for x∈R, n∈N.

⟹^{Derivation}

[∫ x cos nx dx]′ = [A + Bx]′ cos nx + [A + Bx] cos′ nx + [C + Dx]′ sin nx + [C + Dx] sin′ nx
x cos nx = B cos nx − [A + Bx]n sin nx + D sin nx + [C + Dx]n cos nx =

= [B + Cn + Dnx] cos nx + [−An + D − Bnx] sin nx

⟹^{Equations}

0 = B + Cn, x = Dnx, 0 = D − An, 0 = −Bnx ⟹ A = \frac{1}{n^{2}} , B = 0, C = 0, D = \frac{1}{n}

∫ x^{2} cos nx dx = \begin{matrix}u = x^{2}  ⇒ & u′ = 2x \\ v′ = cos nx ⇒ & v = \frac{sin nx}{n} \end{matrix} = \frac{x^{2} sin nx}{n} − \frac{2}{n} ∫ x sin nx dx = \begin{matrix}u = x  ⇒ & u′ = 1 \\ v′ = sin nx ⇒ & v = −\frac{cos nx}{n} \end{matrix} =

= \frac{x^{2} sin nx}{n} − \frac{2}{n} [−\frac{x cos nx}{n} + \frac{1}{n} ∫ cos nx dx] = \frac{x^{2} sin nx}{n} − \frac{2}{n} [−\frac{x cos nx}{n} + \frac{1}{n} \frac{sin nx}{n}] = \frac{2x cos nx}{n^{2}} − \frac{2 sin nx}{n^{3}} + \frac{x^{2} sin nx}{n} + c,

for x∈R, n∈N.

∫ x^{2} cos nx dx = [A+Bx+Cx^{2}] cos nx + [D+Ex+F x^{2}] sin nx + c = \frac{2x}{n^{2}} cos nx + [−\frac{2}{n^{3}} + \frac{x^{2}}{n} ] sin nx + c, for x∈R, n∈N.

⟹^{Derivation}

[∫ x^{2} cos nx dx]′ = [B + 2Cx] cos nx − [A + Bx + Cx^{2}]n sin nx + [E + 2F x] sin nx + [D + Ex + F x^{2}]n cos nx
x^{2} cos nx = [B + Dn + (2C + En)x + F nx^{2}] cos nx + [−An + E + (−Bn + 2F)x − Cnx^{2}] sin nx

⟹^{Equations} 0 = B+Dn, 0 = (2C+En)x, x^{2} = F nx^{2}, 0 = E−An, 0 = (2F −Bn)x, 0 = −Cnx^{2}

⟹ A = 0, B = \frac{2}{n^{2}} , C = 0, D = −\frac{2}{n^{3}} , E = 0, F = \frac{1}{n}

∫ x^{3} cos nx dx = [A+Bx+Cx^{2}+Dx^{3}] cos nx + [E+F x+Gx^{2}+Hx^{3}] sin nx + c = [−\frac{6}{n^{4}} + \frac{3x^{2}}{n^{2}} ] cos nx + [−\frac{6x}{n^{3}} + \frac{x^{3}}{n} ] sin nx + c,

for x∈R, n∈N.

⟹^{Derivation}

[∫ x^{3} cos nx dx]′ = [B + 2Cx + 3Dx^{2}] cos nx − [A + Bx + Cx^{2} + Dx^{3}]n sin nx+

+[F + 2Gx + 3Hx^{2}] sin nx + [E + F x + Gx^{2} + Hx^{3}]n cos nx

x^{3} cos nx = [B + En + (2C + F n)x − (3D + Gn)x^{2} + Hnx^{3}] cos nx + [−An + F + (−Bn + 2G)x + (−Cn + 3H)x^{2} − Dnx^{3}] sin nx

⟹^{Equations} 0=B+En, 0=(2C+F n)x, 0=(3D+Gn)x^{2}, x^{3} =Hnx^{3}, 0=F −An, 0=(2G−Bn)x, 0=(3H −Cn)x^{2}, 0=−Dnx^{3}

⟹ A = −\frac{6}{n^{4}} , B = 0, C = \frac{3}{n^{2}} , D = 0, E = 0, F = −\frac{6}{n^{3}} , G = 0, H = \frac{1}{n}