Question

Evaluate: $\int \frac{1}{x^2\sqrt{1+x^2}}dx.$

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

Answer 1

The correct option is B $=\frac{-\sqrt{x^2+1}}{x}+c$

Let $x=\tan \theta$ or $dx={\sec }^2\theta d\theta$
$\therefore \int \frac{1}{x^2\sqrt{1+x^2}}dx=\int \frac{{\sec }^2\theta d\theta }{{\tan }^2\theta \sec \theta }$
$=\int cosec\theta \cot \theta d\theta$
$=-cosec\theta +c$
$=\frac{-\sqrt{x^2+1}}{x}+c$