Question

Let $f\left(x\right)={\frac{x}{(1+x^n)}}^{1/n}$ for $n\ge 2$ and $g\left(x\right)=\underset{foccursntimes}{\underset{}{(f\circ f\circ ...\circ f)\left(x\right)}}.$ Then $\int x^{n-2}g\left(x\right)dx$

• A. $\frac{1}{n(n-1)}{(1+n{x}^n)}^{1-\frac{1}{n}}+k$
• B. $\frac{1}{n-1}{(1+n{x}^n)}^{1-\frac{1}{n}}+k$
• C. $\frac{1}{n(n+1)}{(1+n{x}^n)}^{1+\frac{1}{n}}+k$
• D. $\frac{1}{n+1}{(1+n{x}^n)}^{1+\frac{1}{n}}+k$

Answer 1

The correct option is A $\frac{1}{n(n-1)}{(1+n{x}^n)}^{1-\frac{1}{n}}+k$

Given $f\left(x\right)=\frac{x}{{(1+x)}^{\frac{1}{n}}}$ for $n\ge 2$

$\therefore ff\left(x\right)=\frac{f\left(x\right)}{{[1+{f\left(x\right)}^n]}^{\frac{1}{n}}}=\frac{x}{{(1+2x)}^{\frac{1}{n}}}$

and $fff\left(x\right)=\frac{x}{{(1+3x^n)}^{\frac{1}{n}}}$

$\therefore g\left(x\right)=(fofo...of)\left(x\right)=\frac{x}{{\left({1+nx}^n\right)}^{\frac{1}{n}}}$

Let $I=\int x^{n-2}g\left(x\right)dx=\int \frac{x^{n-1}dx}{{(1+{nx}^n)}^{\frac{1}{n}}}$

$=\frac{1}{n^2}\int \frac{n^2{x}^{n-1}dx}{{\left({1+nx}^n\right)}^{\frac{1}{n}}}=\frac{1}{n^2}\int \frac{\frac{d}{dx}(1+{nx}^n)}{{\left({1+nx}^n\right)}^{\frac{1}{n}}}dx$

$\therefore I=\frac{1}{n(n-1)}{\left({1+nx}^n\right)}^{1-\frac{1}{n}}+c$