Список интегралов от рациональных функций

Ниже приведён список интегралов (первообразных функций) от рациональных функций.

<math>\int (ax + b)^n dx = \begin{cases} \frac{(ax + b)^{n+1}}{a(n + 1)}, & n\neq -1 \\ \frac{1}{a}\ln\left|ax + b\right|, & n=-1\end{cases}</math>

<math>\int x(ax + b)^n dx = \begin{cases} \frac{a(n + 1)x - b}{a^2(n + 1)(n + 2)} (ax + b)^{n+1}, & n \not\in \{-1, -2\} \\ \frac{x}{a} - \frac{b}{a^2}\ln\left|ax + b\right|, & n=-1 \\ \frac{b}{a^2(ax + b)} + \frac{1}{a^2}\ln\left|ax + b\right|, & n=-2\end{cases}</math>

<math>\int\frac{x}{(ax + b)^n}dx = \frac{a(1 - n)x - b}{a^2(n - 1)(n - 2)(ax + b)^{n-1}}, \quad n\not\in \{1, 2\}</math>

<math>\int\frac{dx}{x^{2^n} + 1} = \sum_{k=1}^{2^{n-1}} \left \{ \frac{1}{2^{n-1}} \left [ \sin \left(\frac{(2k -1) \pi}{2^n}\right) \operatorname{arctg}\left[\left(x - \cos \left(\frac{(2k -1) \pi}{2^n} \right) \right ) \operatorname{cosec} \left(\frac{(2k -1) \pi}{2^n} \right) \right] \right] - \right.

</math> <math> \quad \left. - \, \frac{1}{2^n} \left [ \cos \left(\frac{(2k -1) \pi}{2^n} \right) \ln \left | x^2 - 2 x \cos \left(\frac{(2k -1) \pi}{2^n} \right) + 1 \right | \right ] \right \} + C </math>

<math>\int\frac{x^2}{ax + b}dx = \frac{1}{a^3}\left(\frac{(ax + b)^2}{2} - 2b(ax + b) + b^2\ln\left|ax + b\right|\right)</math>

<math>\int\frac{x^2}{(ax + b)^2}dx = \frac{1}{a^3}\left(ax + b - 2b\ln\left|ax + b\right| - \frac{b^2}{ax + b}\right)</math>

<math>\int\frac{x^2}{(ax + b)^3}dx = \frac{1}{a^3}\left(\ln\left|ax + b\right| + \frac{2b}{ax + b} - \frac{b^2}{2(ax + b)^2}\right)</math>

<math>\int\frac{x^2}{(ax + b)^n}dx = \frac{1}{a^3}\left(-\frac{1}{(n- 3)(ax + b)^{n-3}} + \frac{2b}{(n-2)(ax + b)^{n-2}} - \frac{b^2}{(n - 1)(ax + b)^{n-1}}\right),</math> для <math>n\not\in \{1, 2, 3\}</math>

<math>\int\frac{dx}{x(ax + b)} = -\frac{1}{b}\ln\left|\frac{ax+b}{x}\right|</math>

<math>\int\frac{dx}{x^2(ax+b)} = -\frac{1}{bx} + \frac{a}{b^2}\ln\left|\frac{ax+b}{x}\right|</math>

<math>\int\frac{dx}{x^2(ax+b)^2} = -a\left(\frac{1}{b^2(ax+b)} + \frac{1}{ab^2x} - \frac{2}{b^3}\ln\left|\frac{ax+b}{x}\right|\right)</math>

<math>\int\frac{dx}{a^2x^2+b^2} = \frac{1}{ab}\operatorname{arctg}\frac{ax}{b}</math>

<math>\int\frac{dx}{(x^2+a^2)^2} = \frac{x}{2a^2(x^2+a^2)} + \frac{1}{2a^3}\operatorname{arctg}\frac{x}{a}</math>

<math>\int\frac{dx}{(x^2+a^2)^3} = \frac{x}{4a^2(x^2+a^2)^2} + \frac{3x}{8a^4(x^2+a^2)} + \frac{3}{8a^5}\operatorname{arctg}\frac{x}{a}</math>

<math>\int\frac{dx}{x^2-a^2} = -\frac{1}{a}\,\operatorname{arth}\frac{x}{a} = \frac{1}{2a}\ln\frac{a-x}{a+x},</math> для <math>|x| < |a|</math>

<math>\int\frac{dx}{x^2-a^2} = -\frac{1}{a}\,\operatorname{arcth}\frac{x}{a} = \frac{1}{2a}\ln\frac{x-a}{x+a},</math> для <math>|x| > |a|</math>

<math>\int\frac{dx}{ax^2+bx+c} = \frac{2}{\sqrt{4ac-b^2}}\operatorname{arctg}\frac{2ax+b}{\sqrt{4ac-b^2}},</math> для <math>4ac-b^2>0</math>

<math>\int\frac{dx}{ax^2+bx+c} = -\frac{2}{\sqrt{b^2-4ac}}\,\operatorname{arth}\frac{2ax+b}{\sqrt{b^2-4ac}} = \frac{1}{\sqrt{b^2-4ac}}\ln\left|\frac{2ax+b-\sqrt{b^2-4ac}}{2ax+b+\sqrt{b^2-4ac}}\right|,</math> для <math>4ac-b^2<0</math>

<math>\int\frac{dx}{ax^2+bx+c} = -\frac{2}{2ax+b}\qquad\mbox{(for }4ac-b^2=0\mbox{)}</math>

<math>\int\frac{x}{ax^2+bx+c}dx = \frac{1}{2a}\ln\left|ax^2+bx+c\right|-\frac{b}{2a}\int\frac{dx}{ax^2+bx+c}</math>

<math>\int\frac{mx+n}{ax^2+bx+c}dx = \frac{m}{2a}\ln\left|ax^2+bx+c\right|+\frac{2an-bm}{a\sqrt{4ac-b^2}}\operatorname{arctg}\frac{2ax+b}{\sqrt{4ac-b^2}},</math> для <math>4ac-b^2>0</math>

<math>\int\frac{mx+n}{ax^2+bx+c}dx = \frac{m}{2a}\ln\left|ax^2+bx+c\right|-\frac{2an-bm}{a\sqrt{b^2-4ac}}\,\operatorname{arth}\frac{2ax+b}{\sqrt{b^2-4ac}},</math> для <math>4ac-b^2<0</math>

<math>\int\frac{mx+n}{ax^2+bx+c}dx = \frac{m}{2a}\ln\left|ax^2+bx+c\right|-\frac{2an-bm}{a(2ax+b)},</math> для <math>4ac-b^2=0</math>

<math>\int\frac{dx}{(ax^2+bx+c)^n} = \frac{2ax+b}{(n-1)(4ac-b^2)(ax^2+bx+c)^{n-1}}+\frac{(2n-3)2a}{(n-1)(4ac-b^2)}\int\frac{dx}{(ax^2+bx+c)^{n-1}}</math>

<math>\int\frac{x}{(ax^2+bx+c)^n}dx = \frac{bx+2c}{(n-1)(4ac-b^2)(ax^2+bx+c)^{n-1}}-\frac{b(2n-3)}{(n-1)(4ac-b^2)}\int\frac{dx}{(ax^2+bx+c)^{n-1}}</math>

<math>\int\frac{dx}{x(ax^2+bx+c)} = \frac{1}{2c}\ln\left|\frac{x^2}{ax^2+bx+c}\right|-\frac{b}{2c}\int\frac{dx}{ax^2+bx+c}</math>