Question

$\int \frac{1}{x\sqrt{x^2-1}}dx$ is equal to :

• A. ${\cos }^{-1}x+C$
• B. ${\sec }^{-1}x+C$
• C. ${\cot }^{-1}x+C$
• D. ${\tan }^{-1}x+C$

Answer 1

The correct option is B ${\sec }^{-1}x+C$

Let $I=\int \frac{1}{x\sqrt{x^2-1}}dx$

Put $x=\sec \theta ⟹dx=\sec \theta \tan \theta d\theta$

$\therefore I=\int \frac{\sec \theta \tan \theta }{\sec \theta \sqrt{{\sec }^2\theta -1}}d\theta$

$=\int \frac{\sec \theta \tan \theta }{\sec \theta \tan \theta }d\theta$

$=\int d\theta$

$=\theta +C$

$={\sec }^{-1}x+C$

Hence, $\int \frac{1}{x\sqrt{x^2-1}}dx={\sec }^{-1}x+C$