Question

The value of $\int \frac{1}{x^2\sqrt{1+x^2}}dx$ is

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

Answer 1

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

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