$\int f^{\mathrm{\prime }}(ax+b)\left[f\right(ax+b){]}^{n}dx$ is equal to :

Let $\mathrm{I}=\int {\mathrm{f}}^{\mathrm{\prime }}(\mathrm{ax}+\mathrm{b})\left[\mathrm{f}\right(\mathrm{ax}+\mathrm{b}){]}^{\mathrm{n}}\mathrm{dx}$

Let f(ax + b) =t 

Differentiate both side w.r.t 'x', we get 

      f'(ax + b). a dx = dt

$\begin{array}{l}\therefore \mathrm{I}=\int \frac{1}{\mathrm{a}}[\mathrm{t}{]}^{\mathrm{n}}\mathrm{dt}=\frac{1}{\mathrm{a}}\frac{{\mathrm{t}}^{\mathrm{n}+1}}{\mathrm{n}+1}+\mathrm{C}\\ =\frac{1}{\mathrm{a}(\mathrm{n}+1)}\left[\mathrm{f}\right(\mathrm{ax}+\mathrm{b}){]}^{\mathrm{n}+1}+\mathrm{C}.\end{array}$

       For every $n\ne -1$