Question

Prove that $\int \frac{f^{\prime }\left(x\right)}{\sqrt[n]{nf\left(x\right)}}dx=\frac{\left[f\right(x){]}^{1-\frac{1}{n}}}{1-\frac{1}{n}}+C,n\ne 1$

Answer 1

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

Let $f\left(x\right)=t;f^{\prime }\left(x\right)dx=dt$

$=\int \frac{1}{\sqrt[n]{nt}}dt$

$=\int t^{-\frac{1}{n}}dt$

$=\frac{t^{-\frac{1}{n}+1}}{-\frac{1}{n}+1}+c$

$=\frac{\left[f\right(x){]}^{1-\frac{1}{n}}}{1-\frac{1}{n}}+c$.

$\therefore \int \frac{f^{\prime }\left(x\right)}{\sqrt[n]{nf\left(x\right)}}dx=\frac{\left[f\right(x){]}^{1-\frac{1}{n}}}{1-\frac{1}{n}}+C$