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\int (ax + b)^n \, dx= \frac{(ax + b)^{n+1}}{a(n + 1)} + C \qquad\text{(for } n\neq -1\mbox{)}\,\! (Cavalieri's quadrature formula)
\int\frac{c}{ax + b} \, dx= \frac{c}{a}\ln\left|ax + b\right| + C
\int x(ax + b)^n \, dx= \frac{a(n + 1)x - b}{a^2(n + 1)(n + 2)} (ax + b)^{n+1} + C \qquad\text{(for }n \not\in \{-1, -2\}\mbox{)}
\int\frac{x}{ax + b} \, dx= \frac{x}{a} - \frac{b}{a^2}\ln\left|ax + b\right| + C
\int\frac{x}{(ax + b)^2} \, dx= \frac{b}{a^2(ax + b)} + \frac{1}{a^2}\ln\left|ax + b\right| + C
\int\frac{x}{(ax + b)^n} \, dx= \frac{a(1 - n)x - b}{a^2(n - 1)(n - 2)(ax + b)^{n-1}} + C \qquad\text{(for } n\not\in \{1, 2\}\mbox{)}
\int\frac{f'(x)}{f(x)} \, dx= \ln\left|f(x)\right| + C
\int\frac{x^2}{ax + b} \, dx= \frac{b^2\ln(\left|ax + b\right|)}{a^3}+\frac{ax^2 - 2bx}{2a^2} + C
\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
\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
\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 \qquad\text{(for } n\not\in \{1, 2, 3\}\mbox{)}
\int\frac{1}{x(ax + b)} \, dx = -\frac{1}{b}\ln\left|\frac{ax+b}{x}\right| + C
\int\frac{1}{x^2(ax+b)} \, dx = -\frac{1}{bx} + \frac{a}{b^2}\ln\left|\frac{ax+b}{x}\right| + C
\int\frac{1}{x^2(ax+b)^2} \, dx = -a\left(\frac{1}{b^2(ax+b)} + \frac{1}{ab^2x} - \frac{2}{b^3}\ln\left|\frac{ax+b}{x}\right|\right) + C
\int\frac{1}{x^2+a^2} \, dx = \frac{1}{a}\arctan\frac{x}{a}\,\! + C
\int\frac{1}{x^2-a^2} \, dx = \begin{cases} \displaystyle -\frac{1}{a}\,\mathrm{arctanh}\frac{x}{a} = \frac{1}{2a}\ln\frac{a-x}{a+x} + C & \text{(for }|x| <|a|\mbox{)} \\[12pt] \displaystyle -\frac{1}{a}\,\mathrm{arccoth}\frac{x}{a} = \frac{1}{2a}\ln\frac{x-a}{x+a} + C & \text{(for }|x|> |a| \mbox{)} \end{cases}

For a\neq 0:

\int\frac{1}{ax^2+bx+c} dx = \begin{cases} \displaystyle \frac{2}{\sqrt{4ac-b^2}}\arctan\frac{2ax+b}{\sqrt{4ac-b^2}} + C & \text{(for }4ac-b^2>0\mbox{)} \\[12pt] \displaystyle -\frac{2}{\sqrt{b^2-4ac}}\,\mathrm{arctanh}\frac{2ax+b}{\sqrt{b^2-4ac}} + C = \frac{1}{\sqrt{b^2-4ac}}\ln\left|\frac{2ax+b-\sqrt{b^2-4ac}}{2ax+b+\sqrt{b^2-4ac}}\right| + C & \text{(for }4ac-b^2<0\mbox{)} \\[12pt] \displaystyle -\frac{2}{2ax+b} + C & \text{(for }4ac-b^2=0\mbox{)} \end{cases}
\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} + C
\int\frac{mx+n}{ax^2+bx+c} \, dx = \begin{cases} \displaystyle \frac{m}{2a}\ln\left|ax^2+bx+c\right|+\frac{2an-bm}{a\sqrt{4ac-b^2}}\arctan\frac{2ax+b}{\sqrt{4ac-b^2}} + C &\text{(for }4ac-b^2>0\mbox{)} \\[12pt] \displaystyle \frac{m}{2a}\ln\left|ax^2+bx+c\right|-\frac{2an-bm}{a\sqrt{b^2-4ac}}\,\mathrm{arctanh}\frac{2ax+b}{\sqrt{b^2-4ac}} + C &\text{(for }4ac-b^2<0\mbox{)} \\[12pt] \displaystyle \frac{m}{2a}\ln\left|ax^2+bx+c\right|-\frac{2an-bm}{a(2ax+b)} + C &\text{(for }4ac-b^2=0\mbox{)}\end{cases}
\int\frac{1}{(ax^2+bx+c)^n} \, dx= \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{1}{(ax^2+bx+c)^{n-1}} \, dx + C
\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{1}{(ax^2+bx+c)^{n-1}} \, dx + C
\int\frac{1}{x(ax^2+bx+c)} \, dx= \frac{1}{2c}\ln\left|\frac{x^2}{ax^2+bx+c}\right|-\frac{b}{2c}\int\frac{1}{ax^2+bx+c} \, dx + C
\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 - 2 x \cos \left(\frac{(2k -1) \pi}{2^n} \right) + 1 \right | \right ] \right \} + C