Question

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

• 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 $f\left(x\right)=\frac{x}{(1+x^n{)}^{1/n}}$ for $n\ge 2$
$\therefore f\circ f\left(x\right)=f\left[f\right(x\left)\right]=f\left[\frac{x}{(1+x^n{)}^{1/n}}\right]=\frac{\frac{x}{(1+x^n{)}^{1/n}}}{{[1+\frac{x^n}{1+x^n}]}^{1/n}}=\frac{x}{(1+2x^n{)}^{1/n}}$
Further $f\circ f\circ f\left(x\right)=\frac{x}{(1+3x^n{)}^{1/n}}$
Proceeding in the similar manner. we get
$g\left(x\right)=f\circ f\circ f....\circ f\left(x\right)=\frac{x}{(1+n{x}^n{)}^{1/n}}$
(f occurs n times)
Now, $\int x^{n-2}g\left(x\right)dx=\int \frac{x^{n-1}}{(1+n{x}^n{)}^{1/n}}dx$
Let $1+n{x}^n=t\Rightarrow n^2{x}^{n-1}dx=dt$
$\therefore$ Integral becomes $=\frac{1}{n^2}\int t^{-1/n}dt=\frac{1}{n^2}.\frac{t^{-\frac{1}{n}+1}}{-\frac{1}{n}+1}$
$=\frac{1}{n},\frac{t^{1-1/n}}{n-1}+K=\frac{(1+n{x}^n{)}^{1-1/n}}{n(n-1)}+K$