# How do you find the integral of 5/[sqrt(9x^2-16)] dx?

Apr 1, 2018

The answer is $= \frac{5}{3} \ln \left(| \frac{3}{4} x + \sqrt{{\left(\frac{3}{4} x\right)}^{2} - 1} |\right) + C$

#### Explanation:

Perform some simplification

$\frac{5}{\sqrt{9 {x}^{2} - 16}} = \frac{5}{4 \sqrt{\left({\left(\frac{3}{4} x\right)}^{2}\right) - 1}}$

Let $\frac{3}{4} x = \sec u$, $\implies$, $\frac{3}{4} \mathrm{dx} = \sec u \tan u \mathrm{du}$

$\sqrt{{\left(\frac{3}{4} x\right)}^{2} - 1} = \sqrt{{\sec}^{2} u - 1} = \tan u$

Therefore, the integral is

$I = \int \frac{5 \mathrm{dx}}{\sqrt{9 {x}^{2} - 16}} = \frac{5}{4} \int \frac{\frac{4}{3} \sec u \tan u \mathrm{du}}{\tan u}$

$= \frac{5}{3} \int \sec u \mathrm{du}$

$= \frac{5}{3} \int \frac{\sec u \left(\sec u + \tan u\right) \mathrm{du}}{\sec u + \tan u}$

Let $v = \sec u + \tan u$, $\implies$, $\mathrm{dv} = \left({\sec}^{2} u + \sec u \tan u\right) \mathrm{du}$

Therefore,

$I = \frac{5}{3} \int \frac{\mathrm{dv}}{v}$

$= \frac{5}{3} \ln \left(v\right)$

$= \frac{5}{3} \ln \left(\sec u + \tan u\right)$

$= \frac{5}{3} \ln \left(| \frac{3}{4} x + \sqrt{{\left(\frac{3}{4} x\right)}^{2} - 1} |\right) + C$