Question

If f(x)$=\frac{x}{(1+x^n{)}^{1/n}}$ for n$\ge$ and g (x)= (fofo...of) (x)$\underset{foccursntimes}{\underbrace{(fofo\cdots of)}}(x).$ Then, $\int x^{n-2}g(x)dxequals$

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

Answer 1

Answer: (A)

Given, f(x) $=\frac{x}{(1+x^n{)}^{1/n}}$ for n$\ge$ 2
$\therefore ff(x)=\frac{f(x)}{[1+f(x{)}^n{]}^{1/n}}=\frac{x}{(1+2x^n{)}^{1/n}}$
and fff (x)$=\frac{x}{(1+3x^n{)}^{1/n}}$
$\therefore g(x)=\underset{ntimes}{\underbrace{fofo\cdots of}}(x)=\frac{x}{(1+n{x}^n{)}^{1/n}}$
Let $I=\int x^{n-2}g(x)dx=\int \frac{x^{n-1}dx}{(1+n{x}^n{)}^{1/n}}$
$=\frac{1}{n^2}\int \frac{n^2{x}^{n-1}dx}{(1+n{x}^n{)}^{1/n}}=\frac{1}{n^2}\int \frac{\frac{d}{dx}(1+n{x}^n)}{(1+n{x}^n{)}^{1/n}}dx$
$I=\frac{1}{n(n-1)}(1+n{x}^n{)}^{1-\frac{1}{n}}+c$