fx=x1+xn1n for n≥2and gx=fofofo.....ofx f occurs n times then∫xn−2gxdx=

fof(x)=f(f(x)=f(x)1+(f(x)n1n=x(1+xn)1n1+xn1+xn1n=x(1+2xn)1nfofof(x)=f(fof(x)=x(1+2xn)1n 1+nxn=t⇒n.nxn−1dx=dt∫xn−2g(x)dx=∫xn−1(1+nxn)dx=1n2 ∫dtt1n=1n2t−1n+1(−1n+1)+c=1n2t−1n+1(−1n+1)+c=1n2t−1n+1(−1n+1)+c=1n2n−1+n+c=1n2n−1+nt−1n+c=1n(n−1)1+nxn1−1n+c

fx=x1+xn1n for n≥2and gx=fofofo.....ofx f occurs n times then∫xn−2gxdx=

fof(x)=f(f(x)=f(x)1+(f(x)n1n=x(1+xn)1n1+xn1+xn1n=x(1+2xn)1nfofof(x)=f(fof(x)=x(1+2xn)1n 1+nxn=t⇒n.nxn−1dx=dt∫xn−2g(x)dx=∫xn−1(1+nxn)dx=1n2 ∫dtt1n=1n2t−1n+1(−1n+1)+c=1n2t−1n+1(−1n+1)+c=1n2t−1n+1(−1n+1)+c=1n2n−1+n+c=1n2n−1+nt−1n+c=1n(n−1)1+nxn1−1n+c

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$f (x) = \frac{x}{{(1 + x^{n})}^{\frac{1}{n}}}$ for $n \geq 2$ and $\begin{array}{l} g (x) = (f o f o f o ..... o f) (x) \\ f o c c u r s n t i m e s \end{array}$
then $\int x^{n - 2} g (x) d x =$

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$\frac{1}{n (n - 1)} {(1 + n x^{n})}^{1 - \frac{1}{n}} + k$

$\frac{1}{n - 1} {(1 + n x^{n})}^{1 - \frac{1}{n}} + k$

$\frac{1}{n (n + 1)} {(1 + n x^{x})}^{1 + \frac{1}{n}} + k$

$\frac{1}{n + 1} {(1 + n x^{n})}^{1 + \frac{1}{n}} + k$

answer is A.

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Detailed Solution

$\begin{array}{l} f o f (x) = f (f (x) = \frac{f (x)}{{[1 + (f {(x)}^{n}]}^{\frac{1}{n}}} \\ = \frac{\frac{x}{{(1 + x^{n})}^{\frac{1}{n}}}}{{[1 + \frac{x^{n}}{1 + x^{n}}]}^{\frac{1}{n}}} = \frac{x}{{(1 + 2 x^{n})}^{\frac{1}{n}}} \\ f o f o f (x) = f (f o f (x) = \frac{x}{{(1 + 2 x^{n})}^{\frac{1}{n}}} \\ 1 + n x^{n} = t \Rightarrow n . n x^{n - 1} d x = d t \\ \int x^{n - 2} g (x) d x = \int \frac{x^{n - 1}}{(1 + n x^{n})} d x = \frac{1}{n^{2}} \int \frac{d t}{t^{\frac{1}{n}}} \\ = \frac{1}{n^{2}} \frac{t^{- \frac{1}{n} + 1}}{(- \frac{1}{n} + 1)} + c \\ = \frac{1}{n^{2}} \frac{t^{- \frac{1}{n} + 1}}{(- \frac{1}{n} + 1)} + c \\ = \frac{1}{n^{2}} \frac{t^{- \frac{1}{n} + 1}}{(- \frac{1}{n} + 1)} + c \\ = \frac{1}{n^{2}} (\frac{n}{- 1 + n}) + c \\ = \frac{1}{n^{2}} (\frac{n}{- 1 + n}) t^{- \frac{1}{n}} + c \\ = \frac{1}{n (n - 1)} {(1 + n x^{n})}^{1 - \frac{1}{n}} + c \end{array}$