Question

$\int \frac{1}{x\sqrt{1+x^n}}dx=\dots \dots$

• A. $\frac{1}{n}\log \sqrt{\frac{1+x^n}{1-x^n}}+c$
• B. $\frac{1}{n}\log \left|\frac{\sqrt{1+x^n}-1}{\sqrt{1+x^n}+1}\right|+c$
• C. $\frac{1}{n}\log \left|\frac{1+x^n}{1-x^n}\right|+c$
• D. $\frac{1}{n}\log \left|\frac{\sqrt{1+x^n}+1}{\sqrt{1+x^n}-1}\right|+c$

Answer 1

Answer: (B)

$\frac{1}{n}\log \left|\frac{\sqrt{1+x^n}-1}{\sqrt{1+x^n}+1}\right|+c$

$I=\int \frac{1}{x\sqrt{1+x^n}}dx$

Let $1+x^n=t^2\Rightarrow n{x}^{n-1}dx=2tdt$

$\Rightarrow n{x}^n\frac{dx}{x}=2tdt\Rightarrow \frac{dx}{x}=\frac{2tdt}{n(t^2-1)}$

$I=\int \frac{2tdt}{n(t^2-1)t}=\int \frac{1}{n}(\frac{1}{t-1}-\frac{1}{t+1})dt=\frac{1}{n}\ln (t-1)-\ln (t+1)$

$=\frac{1}{n}\ln \left|\frac{\sqrt{1+x^n}-1}{\sqrt{1+x^n}+1}\right|+C$

Hence answer is B