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 \[1+n{{x}^{n}}=t\]\[\Rightarrow \]\[{{x}^{n\,-\,1}}dx=\frac{dt}{{{n}^{2}}}\] |
\[\therefore \]\[I=\frac{1}{{{n}^{2}}}\int{\frac{dt}{{{t}^{1/n}}}}\] |
\[\Rightarrow \]\[I=\frac{1}{{{n}^{2}}}\frac{{{t}^{-\frac{1}{n}+1}}}{-\frac{1}{n}+1}+C\] \[\Rightarrow \] \[I=\frac{1}{n\,(n-1)}{{t}^{1-\frac{1}{n}}}+C\] \[\Rightarrow \] \[I=\frac{1}{n\,(n-1)}{{(1+n{{x}^{n}})}^{1-\frac{1}{n}}}+C\] |
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