Question

The value of the integral $\int \frac{dx}{x\sqrt{1+x^n}}$ is:
(where $c$ is integration constant)

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

Answer 1

The correct option is C $\frac{1}{n}\ln \left|\frac{\sqrt{1+x^n}-1}{\sqrt{1+x^n}+1}\right|+c$

Let $I=\int \frac{dx}{x\sqrt{1+x^n}}$
Put $1+x^n=t^2$
$\Rightarrow n{x}^{n-1}dx=2tdt\Rightarrow n{x}^n\frac{dx}{x}=2tdt\Rightarrow I=\int \frac{2tdt}{n(t^2-1)t}\Rightarrow I=\frac{1}{n}\int \frac{2dt}{t^2-1}\Rightarrow I=\frac{1}{n}\left(\int \frac{dt}{t-1}-\int \frac{dt}{t+1}\right)\Rightarrow I=\frac{1}{n}\ln \left|\frac{t-1}{t+1}\right|+c\Rightarrow I=\frac{1}{n}\ln \left|\frac{\sqrt{1+x^n}-1}{\sqrt{1+x^n}+1}\right|+c$