$∫\frac{dx}{x(x^n+1)}$ is equal to
Answer & explanation
Correct answer: option 2
Let $I=∫\frac{dx}{x(x^n+1)}=∫\frac{dx}{x^{n+1}(1+\frac{1}{x^n})}$.
If $(1+\frac{1}{x^n})=p$, then $-\frac{n}{x^{n+1}}dx=dp$
$⇒I=-\frac{1}{n}∫\frac{dp}{p}=\frac{-1}{n}log_ep+c=-\frac{1}{n}log_e(\frac{x^n+1}{x^n})+c$
Hence (B) is the correct answer.