Let $f\left(x\right)=\frac{x}{{\left(1+x^n\right)}^{1/n}}$ for $n\ge 2$ and $g\left(x\right)=$

$\frac{f\circ f\dots of}{n\text{ times }}\text{.}$Then $\int x^{n-2}g\left(x\right)dx$ equal 

We have 

$\begin{array}{l}(f\circ f)\left(x\right)=f\left(f\right(x\left)\right)=\frac{f\left(x\right)}{{\left[1+\left(f\right(x){)}^n\right]}^{1/n}}\\ =\frac{x}{{\left(1+2x^n\right)}^{1/n}}\end{array}$

Similarly, $\left(\text{ fofof }\right)\left(x\right)=\frac{x}{{\left(1+3x^n\right)}^{1/n}}$

Continuing in this way we get 

$g\left(x\right)=\underset{n\text{ times }}{\underbrace{(\text{ fof }\cdots \text{ of })\left(x\right)}}=\frac{x}{{\left(1+n{x}^n\right)}^{1/n}}$

Thus , $I=\int x^{n-2}\frac{x}{{\left(1+n{x}^n\right)}^{1/n}}dx$

Put $1+n{x}^n=t^n$ to obtain 

$\begin{array}{l}I=\frac{1}{n}\int \frac{t^{n-1}}{t}dt=\frac{1}{n}\int t^{n-2}dt\\ =\frac{1}{n(n-1)}{t}^{n-1}+C=\frac{1}{n(n-1)}{\left(1+n{x}^n\right)}^{1-1/n}+C\end{array}$