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${\displaystyle \int (ax+b{)}^{n}dx=\frac{(ax+b{)}^{n+1}}{a(n+1)}+C\text{(for }n\ne -1\text{)}}$

(Cavalieri's quadrature formula)

${\displaystyle \int \frac{c}{ax+b}dx=\frac{c}{a}\ln \left|ax+b\right|+C}$

${\displaystyle \int x(ax+b{)}^{n}dx=\frac{a(n+1)x-b}{a^2(n+1)(n+2)}(ax+b{)}^{n+1}+C\text{(for }n\notin \{-1,-2\}\text{)}}$

${\displaystyle \int \frac{x}{ax+b}dx=\frac{x}{a}-\frac{b}{a^2}\ln \left|ax+b\right|+C}$

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

${\displaystyle \int \frac{x}{(ax+b{)}^n}dx=\frac{a(1-n)x-b}{a^2(n-1)(n-2)(ax+b{)}^{n-1}}+C\text{(for }n\notin \{1,2\}\text{)}}$

${\displaystyle \int \frac{f^{\prime }(x)}{f(x)}dx=\ln \left|f(x)\right|+C}$

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

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

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

${\displaystyle \int \frac{x^2}{(ax+b{)}^n}dx=\frac{1}{a^3}\left(-\frac{(ax+b{)}^{3-n}}{(n-3)}+\frac{2b(ax+b{)}^{2-n}}{(n-2)}-\frac{b^2(ax+b{)}^{1-n}}{(n-1)}\right)+C\text{(for }n\notin \{1,2,3\}\text{)}}$

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

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

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

${\displaystyle \int \frac{1}{x^2+a^2}dx=\frac{1}{a}\arctan \frac{x}{a}+C}$

For ${\displaystyle a\ne 0:}$

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

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

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

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

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