Question

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

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

Answer 1

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

$\int \frac{x^2-2}{x^3\sqrt{x^2-1}}dx=\int \frac{dx}{x\sqrt{x^2-1}}-2\int \frac{dx}{x^3\sqrt{x^2-1}}=se{c}^{-1}x-2\int \frac{sec\theta tan\theta }{se{c}^3\theta tan\theta }d\theta =[putx=sec\theta \Rightarrow dx=sec\theta tan\theta d\theta ]=se{c}^{-1}x-2\int co{s}^2\theta d\theta =se{c}^{-1}x-2\int (1+cos2\theta )d\theta =se{c}^{-1}x-(\theta +\frac{sin2\theta }{2})+C=se{c}^{-1}x-se{c}^{-1}x-\frac{\sqrt{x^2-1}}{x^2}+C=-\frac{\sqrt{x^2-1}}{x^2}+C$