Question

Prove the following:
$\int \left[f\right(x\left){]}^n{f}^{\prime }\right(x)dx=\frac{\left[f\right(x){]}^{n+1}}{n+1}+C$, if $n\ne -1$

Answer 1

$I=\int \left[f\right(x\left){]}^n{f}^{\prime }\right(x)dx$.

Let $f\left(x\right)=t,f^{\prime }\left(x\right)dx=dt$
$I=\int t^{n}dt$
$=\frac{t^{n+1}}{n+1}+c$
$=\frac{\left[f\right(x){]}^{n+1}}{n+1}+c$