question_answer

Let \[f\,(x)=\frac{x}{{{(1+{{x}^{n}})}^{1/n}}}\] for \[n\ge 2\] and \[g(x)=\underbrace{({{f}_{0}}{{f}_{0}}...f)}_{n\,\,\text{times}}f(x).\] Then, \[\int{{{x}^{n-2}}g(x)\,dx}\] is equal to

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) \[\frac{1}{n+1}{{(1+n{{x}^{n}})}^{1+\frac{1}{n}}}+C\]

Correct Answer: A

Solution :

[a]

We have, \[f(x)=\frac{x}{{{(1+{{x}^{n}})}^{\frac{1}{n}}}}\]

\[f\,(f(x)=\frac{f(x)}{1+f{{(x)}^{n}}{{)}^{\frac{1}{n}}}}=\frac{x}{{{(1+2{{x}^{n}})}^{\frac{1}{n}}}}\]

\[\Rightarrow \]\[I=\int{{{x}^{n\,-\,2}}g\,(x)\,dx}\]\[=\int{{{x}^{n\,-\,2}}\frac{x}{{{(1+n{{x}^{n}})}^{\frac{1}{n}}}}dx}\]\[=\int{\frac{{{x}^{n\,-\,1}}}{{{(1+n{{x}^{n}})}^{\frac{1}{n}}}}dx}\]

Put \[1+n{{x}^{n}}=t\]\[\Rightarrow \]\[{{x}^{n\,-\,1}}dx=\frac{dt}{{{n}^{2}}}\]

\[\therefore \]\[I=\frac{1}{{{n}^{2}}}\int{\frac{dt}{{{t}^{1/n}}}}\]

\[\Rightarrow \]\[I=\frac{1}{{{n}^{2}}}\frac{{{t}^{-\frac{1}{n}+1}}}{-\frac{1}{n}+1}+C\] \[\Rightarrow \] \[I=\frac{1}{n\,(n-1)}{{t}^{1-\frac{1}{n}}}+C\] \[\Rightarrow \] \[I=\frac{1}{n\,(n-1)}{{(1+n{{x}^{n}})}^{1-\frac{1}{n}}}+C\]