$\int (ax+b)^n ~dx = \dfrac{(ax+b)^{n+1}}{a(n+1)}, \quad (\text{for } n \ne 1)\\$ $\int \dfrac{1}{ax+b}~dx = \dfrac{1}{a}~\ln|ax+b|\\$ $\int x (ax+b)^n~dx = \dfrac{a(n+1)x-b}{a^2(n+1)(n+2)}~(ax+b)^{n+1}, \quad (\text{for } n \ne -1, n\ne-2)\\$ $\int \dfrac{x}{ax+b}~dx = \dfrac{x}{2} - \dfrac{b}{a^2}~\ln|ax+b|\\$ $\int \dfrac{x}{(ax+b)^2}~dx = \dfrac{b}{a^2(ax+b)} - \dfrac{1}{a^2}~\ln|ax+b|\\$ $\int \dfrac{x^2}{ax+b} ~dx = \dfrac{1}{a^3} \left(\dfrac{(ax+b)^2}{2}-2b(ax+b)+b^2~\ln|ax+b| \right)\\$ $\int \dfrac{x^2}{(ax+b)^2} ~dx = \dfrac{1}{a^3} \left(ax+b-2\,b\,\ln|ax+b| - \dfrac{b^2}{ax+b}\right)\\$ $\int \dfrac{x^2}{(ax+b)^3} ~dx = \dfrac{1}{a^3} \left( \ln|ax+b| + \dfrac{2b}{ax+b} - \dfrac{b^2}{2(ax+b)^2} \right)\\$ $\int \dfrac{x^2}{(ax+b)^n} ~dx =                  \dfrac{1}{a^3} \left( -\dfrac{(ax+b)^{3-n}}{n-3} + \dfrac{2b(a+b)^{2-n}}{n-2} - \dfrac{b^2(ax+b)^{1-n}}{n-1}\right)\\$ $\int \dfrac{1}{x(ax+b)}~dx = -\dfrac{1}{b}~\ln\left|\dfrac{ax+b}{x}\right|\\$ $\int \dfrac{1}{x^2(ax+b)^2}~dx = -\dfrac{1}{bx} + \dfrac{a}{b^2}~ \ln\left|\dfrac{ax+b}{x}\right|\\$ $\int \dfrac{1}{x^2(ax+b)^2}~dx = -a\left(\dfrac{1}{b^2(ax+b)} + \dfrac{1}{ab^2x} - \dfrac{2}{b^3}~\ln\left|\dfrac{ax+b}{x} \right|\right)\\$ $\dfrac{1}{x^2+a^2} ~dx = \dfrac{1}{a}~\tan^{-1}\dfrac{x}{a}\\$ $\dfrac{1}{x^2 - a^2} ~dx = \dfrac{1}{2a} ~\ln\left| \dfrac{x-a}{x+a}\right|\\$ $\int \dfrac{1}{ax^2 + bx + c} ~dx = \left\{                 \begin{aligned}                 & \frac{2}{\sqrt{4ac - b^2}} \tan^{-1} \frac{2ax + b}{\sqrt{4ac-b^2}} \quad  \text{for } 4ac - b^2 > 0 \\                 & \frac{2}{\sqrt{b^2 - 4ac}}~ \ln \left| \frac{2ax + b - \sqrt{b^2-4ac}}{2ax + b + \sqrt{b^2-4ac}} \right| \quad \text{for } 4ac - b^2 < 0 \\                 & -\dfrac{2}{2ax+b}  \quad  \text{for } 4ac - b^2 = 0                 \end{aligned} \right.\\$ $\int \dfrac{x}{ax^2 + bx +c} ~dx = \dfrac{1}{2a}~ \ln\left|ax^2+bx+c\right| - \dfrac{b}{2a}\int \dfrac{~dx}{ax^2+bx+c}\\$ $\int \dfrac{1}{(ax^2+bx+c)^n} ~dx = \dfrac{2ax+b}{(n-1)(4ac-b^2)(ax+bx+c)^{n-1}}+                  \dfrac{2(2n-3)a}{(n-1)(4ac - b^2)}\int\dfrac{~dx}{(ax^2+bx+c)^{n-1}}\\$ $\int \dfrac{1}{x(ax^2 + bx + c)} ~dx = \dfrac{1}{2c}~\ln\left|\dfrac{x^2}{ax^2+bx+c}\right| -                 \dfrac{b}{2c} \int \dfrac{1}{ax^2+bx+c}~dx$

---