Question

$\int \frac{1}{x^2\sqrt{1+x^2}}dx$ is equal to (where $C$ is integration constant)

• A. $-\frac{\sqrt{x^2+1}}{x}+C$
• B. $\frac{\sqrt{x^2+1}}{x}+C$
• C. $\frac{\sqrt{x^2+1}}{x^2}+C$
• D. $-\frac{\sqrt{x^2+1}}{x^2}+C$

Answer 1

The correct option is A $-\frac{\sqrt{x^2+1}}{x}+C$

Let $x=\tan \theta$
$\Rightarrow dx={\sec }^2\theta d\theta$
$\Rightarrow \int \frac{1}{x^2\sqrt{1+x^2}}dx$
$=\int \frac{{\sec }^2\theta d\theta }{{\tan }^2\theta \sec \theta }$
$=\int \text{cosec}\theta \cot \theta d\theta$
$=-\text{cosec}\theta +C$
$=-\frac{\sqrt{x^2+1}}{x}+C$