Question

If f(x)$= \frac {x}{(1+x^n)^{1/n}}$ for n$\ge$ and g (x)= (fofo...of) (x)$\underbrace { ( fo fo \cdots of)}_{ f \, occurs \, n \, times } (x).$ Then, $\int \limits x^{n-2}g (x)dx equals $

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

Answer 1

Answer: (A)

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