A) \[1+{{x}^{n}}\]
B) \[1+{{x}^{-n}}\]
C) \[{{x}^{n}}+{{x}^{-n}}\]
D) None of these
Correct Answer: B
Solution :
We have; \[\int{\frac{dx}{{{x}^{2}}{{({{x}^{n}}+1)}^{\frac{(n-1)}{n}}}}=\int{\frac{dx}{{{x}^{2}}\cdot {{x}^{n-1}}{{\left( 1+\frac{1}{{{x}^{n}}} \right)}^{\frac{(n-1)}{n}}}}}}\] \[=\int{\frac{dx}{{{x}^{n+1}}{{(1+{{x}^{-n}})}^{\frac{(n-1)}{n}}}}}\] Put, \[1+{{x}^{-n}}=t\] \[\therefore \] \[-n{{x}^{-n-1}}dx=dt\] \[\Rightarrow \] \[\frac{dx}{{{x}^{n+1}}}=\frac{-dt}{n}\] \[\Rightarrow \] \[\int{\frac{dx}{{{x}^{2}}{{({{x}^{n}}+1)}^{\frac{(n-1)}{n}}}}=-\frac{1}{n}\int{\frac{dt}{{{t}^{\frac{(n-1)}{n}}}}}}\] \[=-\frac{1}{n}\int{{{t}^{(t/n)-1}}}dt=\frac{1}{n}\cdot \frac{{{t}^{\left( \frac{1}{n}-1+1 \right)}}}{\left( \frac{1}{n}-1+1 \right)}+c\] \[=-{{t}^{1/n}}+c=-{{(1+{{x}^{-n}})}^{\frac{1}{n}}}+c\]
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