Tire value of ∫1xn1+xn′1/ndx,(n∈N), is

LetI=∫1xn1+xn1/ndx=∫1xn+11+1xn1/ndx⇒ I=∫1+x−n−1/nx−n−1dx⇒ l=−1n∫1+x−n−1/nd1+x−n⇒ l=−1n1+x−n−1+1−1n+1+C=11−n1+1xn1−1π+C

Tire value of ∫1xn1+xn′1/ndx,(n∈N), is

LetI=∫1xn1+xn1/ndx=∫1xn+11+1xn1/ndx⇒ I=∫1+x−n−1/nx−n−1dx⇒ l=−1n∫1+x−n−1/nd1+x−n⇒ l=−1n1+x−n−1+1−1n+1+C=11−n1+1xn1−1π+C

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Tire value of $\int \frac{1}{x^{n} {(1 + x^{n'})}^{1 / n}} d x, (n \in N)$ , is

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$- \frac{1}{1 + n} {\{1 + \frac{1}{x^{n}}\}}^{1 - \frac{1}{n}} + C$

$\frac{1}{1 - n} {\{1 + \frac{1}{x^{n}}\}}^{1 - \frac{1}{n}} + C$

$- \frac{1}{1 - n} {\{1 - \frac{1}{x^{n}}\}}^{1 - \frac{1}{n}} + C$

$\frac{1}{1 + n} {\{1 - \frac{1}{x^{n}}\}}^{1 - \frac{1}{H}} + C$

answer is A.

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Detailed Solution

Let

$\begin{array}{l} I = \int \frac{1}{x^{n} {(1 + x^{n})}^{1 / n}} d x = \int \frac{1}{x^{n + 1} {(1 + \frac{1}{x^{n}})}^{1 / n} d x} \\ \Rightarrow I = \int {(1 + x^{- n})}^{- 1 / n} x^{- n - 1} d x \\ \Rightarrow l = - \frac{1}{n} \int {(1 + x^{- n})}^{- 1 / n} d (1 + x^{- n}) \\ \Rightarrow l = - \frac{1}{n} \frac{{(1 + x^{- n})}^{- 1} + 1}{(- \frac{1}{n} + 1)} + C = \frac{1}{1 - n} {\{1 + \frac{1}{x^{n}}\}}^{1 - \frac{1}{π}} + C \end{array}$