Question

$\int \frac{x^2-2}{x^3\sqrt{x^2-1}}dx$ is equal to.

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

Answer 1

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

Let $I=\int \frac{x^2-2}{x^3\sqrt{x^2-1}}dx$

$=\int \frac{x^2}{x^3\sqrt{x^2-1}}dx-2\int \frac{1}{x^3\sqrt{x^2-1}}dx$

$=\int \frac{1}{x\sqrt{x^2-1}}dx-2\int \frac{1}{x^3\sqrt{x^2-1}}dx$

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

$\therefore I=\int \frac{1}{x\sqrt{x^2-1}}dx-2\int \frac{\sec \theta \tan \theta }{{\sec }^3\theta \tan \theta }d\theta$

$I=\int \frac{1}{x\sqrt{x^2-1}}dx-2\int {\cos }^2\theta d\theta$

$={\sec }^{-1}x-\int (1+\cos 2\theta )d\theta$ ..... $[\because \cos 2\theta =2{\cos }^2\theta -1]$

$={\sec }^{-1}x-(\theta +\frac{\sin 2\theta }{2})+C$

$={\sec }^{-1}x-{\sec }^{-1}x-\frac{\sqrt{x^2-1}}{x^2}+C$

$\therefore I=-\frac{\sqrt{x^2-1}}{x^2}+C$.