Let \(f(x)=\frac{x}{\left(1+x^n\right)^{1 / n}}\) for \(n \geq 2\) and \(g(x)=\underbrace{(f \circ f o \ldots o f)}_{f \text { occurs } n \text { times }}(x)\). Then, \(\int x^{n-2} g(x) d x\) equals

Here, \(\begin{array}{rlrl} & \text { and } & f f(x) & =\frac{f(x)}{\left[1+f(x)^n\right]^{1 / n}}=\frac{x}{\left(1+2 x^n\right)^{1 / n}} \\ & \therefore \quad f f f(x) & =\frac{x}{\left(1+3 x^n\right)^{1 / n}} \\ & \text { Let } & g(x) & =\underbrace{(f \circ f o \ldots o f)}_{n \text { times }}(x)=\frac{x}{\left(1+n x^n\right)^{1 / n}} \\ I & =\int x^{n-2} g(x) d x=\int \frac{x^{n-1} d x}{\left(1+n x^n\right)^{1 / n}} \\ & & =\frac{1}{n^2} \int \frac{n^2 x^{n-1} d x}{\left(1+n x^n\right)^{1 / n}}=\frac{1}{n^2} \int \frac{\frac{d}{d x}\left(1+n x^n\right)}{\left(1+n x^n\right)^{1 / n}} d x\end{array}\)
\(\therefore \quad I=\frac{1}{n(n-1)}\left(1+n x^n\right)^{1-\frac{1}{n}}+k\).