# How do you evaluate the integral int sqrt(4x^2-1)/x^2?

Apr 30, 2018

$I = 2 \ln | 2 x + \sqrt{4 {x}^{2} - 1} | - \frac{\sqrt{4 {x}^{2} - 1}}{x} + c$

#### Explanation:

Here,

$I = \int \frac{\sqrt{4 {x}^{2} - 1}}{x} ^ 2 \mathrm{dx}$

Let,

$2 x = \sec u \implies \mathrm{dx} = \frac{1}{2} \sec u \tan u \mathrm{du}$

$\therefore 4 {x}^{2} = {\sec}^{2} u \mathmr{and} {x}^{2} = \frac{1}{4} {\sec}^{2} u$

$I = \int \frac{\sqrt{{\sec}^{2} u - 1}}{\frac{1}{4} {\sec}^{2} u} \times \frac{1}{2} \sec u \tan u \mathrm{du}$

$= \frac{4}{2} \int \tan \frac{u}{\sec} ^ 2 u \times \sec u \tan u \mathrm{du}$

$= 2 \int {\tan}^{2} \frac{u}{\sec} u \mathrm{du}$

$= 2 \int \frac{{\sec}^{2} u - 1}{\sec} u \mathrm{du}$

$= 2 \int \left(\sec u - \frac{1}{\sec} u\right) \mathrm{du}$

$= 2 \int \left(\sec u - \cos u\right) \mathrm{du}$

$= 2 \left[\ln | \sec u + \tan u | - \sin u\right] + c$

$= 2 \ln | \sec u + \sqrt{{\sec}^{2} u - 1} | - 2 \left(\sin \frac{u}{\cos} u \times \cos u\right) + c$

$= 2 \ln | \sec u + \sqrt{{\sec}^{2} u - 1} | - 2 \tan u \times \frac{1}{\sec} u + c$

$= 2 \ln | \sec u + \sqrt{{\sec}^{2} u - 1} | - \frac{2 \sqrt{{\sec}^{2} u - 1}}{\sec} u + c$

Subst. back, $\sec u = 2 x$, we get

$I = 2 \ln | 2 x + \sqrt{4 {x}^{2} - 1} | - \frac{2 \sqrt{4 {x}^{2} - 1}}{2 x} + c$

$I = 2 \ln | 2 x + \sqrt{4 {x}^{2} - 1} | - \frac{\sqrt{4 {x}^{2} - 1}}{x} + c$