# How do you evaluate the integral int dx/(x^2sqrt(x^2-3))?

Aug 31, 2017

After using $x = \sqrt{3} \cdot \sec u$ and $\mathrm{dx} = \sqrt{3} \cdot \sec u \cdot \tan u \cdot \mathrm{du}$ transforms, this integral became,

$\int \frac{\sqrt{3} \cdot \sec u \cdot \tan u \cdot \left(\mathrm{du}\right)}{3 {\left(\sec u\right)}^{2} \cdot \sqrt{3} \cdot \tan u}$

=$\frac{1}{3} \cdot \int \frac{\mathrm{du}}{\sec} u$

=$\frac{1}{3} \cdot \int \cos u \cdot \mathrm{du}$

=$\frac{1}{3} \cdot \sin u + C$

After using $x = \sqrt{3} \cdot \sec u$, $\sec u = \frac{x}{\sqrt{3}}$, $\tan u = \frac{\sqrt{{x}^{2} - 3}}{\sqrt{3}}$ and $\sin u = \tan \frac{u}{\sec} u = \frac{\sqrt{{x}^{2} - 3}}{x}$ inverse transforms, I found,

$\int \frac{\mathrm{dx}}{{x}^{2} \cdot \sqrt{{x}^{2} - 3}} = \frac{\sqrt{{x}^{2} - 3}}{3 x} + C$

#### Explanation:

I used $x = \sqrt{3} \cdot \sec u$ and $\mathrm{dx} = \sqrt{3} \cdot \sec u \cdot \tan u \cdot \mathrm{du}$ transform