Список интегралов от рациональных функций

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

$\int (ax+b{)}^{n}dx=\{\begin{array}{cc}\frac{(ax+b{)}^{n+1}}{a(n+1)},& n\ne -1\\ \frac{1}{a}\ln |ax+b|,& n=-1\end{array}$

$\int x(ax+b{)}^{n}dx=\{\begin{array}{cc}\frac{a(n+1)x-b}{a^2(n+1)(n+2)}(ax+b{)}^{n+1},& n\in ̸\{-1,-2\}\\ \frac{x}{a}-\frac{b}{a^2}\ln |ax+b|,& n=-1\\ \frac{b}{a^2(ax+b)}+\frac{1}{a^2}\ln |ax+b|,& n=-2\end{array}$

$\int \frac{x}{(ax+b{)}^n}dx=\frac{a(1-n)x-b}{a^2(n-1)(n-2)(ax+b{)}^{n-1}},n\in ̸\{1,2\}$

$\int \frac{dx}{x^{2^n}+1}=\sum _{k=1}^{2^{n-1}}\{\frac{1}{2^{n-1}}[\sin \left(\frac{(2k-1)\pi }{2^n}\right)\mathrm{arctg}[(x-\cos \left(\frac{(2k-1)\pi }{2^n}\right))\mathrm{cosec}\left(\frac{(2k-1)\pi }{2^n}\right)]]-$ $-\frac{1}{2^n}[\cos \left(\frac{(2k-1)\pi }{2^n}\right)\ln |x^2-2x\cos \left(\frac{(2k-1)\pi }{2^n}\right)+1|]\}+C$

$\int \frac{x^2}{ax+b}dx=\frac{1}{a^3}(\frac{(ax+b{)}^2}{2}-2b(ax+b)+b^2\ln |ax+b|)$

$\int \frac{x^2}{(ax+b{)}^2}dx=\frac{1}{a^3}(ax+b-2b\ln |ax+b|-\frac{b^2}{ax+b})$

$\int \frac{x^2}{(ax+b{)}^3}dx=\frac{1}{a^3}(\ln |ax+b|+\frac{2b}{ax+b}-\frac{b^2}{2(ax+b{)}^2})$

$\int \frac{x^2}{(ax+b{)}^n}dx=\frac{1}{a^3}(-\frac{1}{(n-3)(ax+b{)}^{n-3}}+\frac{2b}{(n-2)(ax+b{)}^{n-2}}-\frac{b^2}{(n-1)(ax+b{)}^{n-1}}),$ для $n\in ̸\{1,2,3\}$

$\int \frac{dx}{x(ax+b)}=-\frac{1}{b}\ln \left|\frac{ax+b}{x}\right|$

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

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

$\int \frac{dx}{a^2{x}^2+b^2}=\frac{1}{ab}\mathrm{arctg}\frac{ax}{b}$

$\int \frac{dx}{(x^2+a^2{)}^2}=\frac{x}{2a^2(x^2+a^2)}+\frac{1}{2a^3}\mathrm{arctg}\frac{x}{a}$

$\int \frac{dx}{(x^2+a^2{)}^3}=\frac{x}{4a^2(x^2+a^2{)}^2}+\frac{3x}{8a^4(x^2+a^2)}+\frac{3}{8a^5}\mathrm{arctg}\frac{x}{a}$

$\int \frac{dx}{x^2-a^2}=-\frac{1}{a}\mathrm{arth}\frac{x}{a}=\frac{1}{2a}\ln \frac{a-x}{a+x},$ для $|x|<|a|$

$\int \frac{dx}{x^2-a^2}=-\frac{1}{a}\mathrm{arcth}\frac{x}{a}=\frac{1}{2a}\ln \frac{x-a}{x+a},$ для $|x|>|a|$

$\int \frac{dx}{a{x}^2+bx+c}=\frac{2}{\sqrt{4ac-b^2}}\mathrm{arctg}\frac{2ax+b}{\sqrt{4ac-b^2}},$ для $4ac-b^2>0$

$\int \frac{dx}{a{x}^2+bx+c}=-\frac{2}{\sqrt{b^2-4ac}}\mathrm{arth}\frac{2ax+b}{\sqrt{b^2-4ac}}=\frac{1}{\sqrt{b^2-4ac}}\ln \left|\frac{2ax+b-\sqrt{b^2-4ac}}{2ax+b+\sqrt{b^2-4ac}}\right|,$ для $4ac-b^2<0$

$\int \frac{dx}{a{x}^2+bx+c}=-\frac{2}{2ax+b}\text{(for }4ac-b^2=0\text{)}$

$\int \frac{x}{a{x}^2+bx+c}dx=\frac{1}{2a}\ln |a{x}^2+bx+c|-\frac{b}{2a}\int \frac{dx}{a{x}^2+bx+c}$

$\int \frac{mx+n}{a{x}^2+bx+c}dx=\frac{m}{2a}\ln |a{x}^2+bx+c|+\frac{2an-bm}{a\sqrt{4ac-b^2}}\mathrm{arctg}\frac{2ax+b}{\sqrt{4ac-b^2}},$ для $4ac-b^2>0$

$\int \frac{mx+n}{a{x}^2+bx+c}dx=\frac{m}{2a}\ln |a{x}^2+bx+c|-\frac{2an-bm}{a\sqrt{b^2-4ac}}\mathrm{arth}\frac{2ax+b}{\sqrt{b^2-4ac}},$ для $4ac-b^2<0$

$\int \frac{mx+n}{a{x}^2+bx+c}dx=\frac{m}{2a}\ln |a{x}^2+bx+c|-\frac{2an-bm}{a(2ax+b)},$ для $4ac-b^2=0$

$\int \frac{dx}{(a{x}^2+bx+c{)}^n}=\frac{2ax+b}{(n-1)(4ac-b^2)(a{x}^2+bx+c{)}^{n-1}}+\frac{(2n-3)2a}{(n-1)(4ac-b^2)}\int \frac{dx}{(a{x}^2+bx+c{)}^{n-1}}$

$\int \frac{x}{(a{x}^2+bx+c{)}^n}dx=\frac{bx+2c}{(n-1)(4ac-b^2)(a{x}^2+bx+c{)}^{n-1}}-\frac{b(2n-3)}{(n-1)(4ac-b^2)}\int \frac{dx}{(a{x}^2+bx+c{)}^{n-1}}$

$\int \frac{dx}{x(a{x}^2+bx+c)}=\frac{1}{2c}\ln \left|\frac{x^2}{a{x}^2+bx+c}\right|-\frac{b}{2c}\int \frac{dx}{a{x}^2+bx+c}$