# How do you find the integral of (2dx) /(x^3sqrt(x^2 - 1))?

Apr 20, 2018

$\int \frac{2 \mathrm{dx}}{{x}^{3} \sqrt{{x}^{2} - 1}} = \arctan \left(\sqrt{{x}^{2} - 1}\right) + \frac{\sqrt{{x}^{2} - 1}}{x} ^ 2 + C$

#### Explanation:

For $x \in \left(1 , + \infty\right)$ substitute:

$x = \sec t$

$\mathrm{dx} = \sec t \tan t \mathrm{dt}$

with $t \in \left(0 , \frac{\pi}{2}\right)$

so:

$\int \frac{2 \mathrm{dx}}{{x}^{3} \sqrt{{x}^{2} - 1}} = 2 \int \frac{\sec t \tan t \mathrm{dt}}{{\sec}^{3} t \sqrt{{\sec}^{2} t - 1}}$

Now:

${\sec}^{2} t - 1 = {\tan}^{2} t$

and as for $t \in \left(0 , \frac{\pi}{2}\right)$ the tangent is positive:

$\sqrt{{\sec}^{2} t - 1} = \tan t$

then:

$\int \frac{2 \mathrm{dx}}{{x}^{3} \sqrt{{x}^{2} - 1}} = 2 \int \frac{\sec t \tan t \mathrm{dt}}{{\sec}^{3} t \tan t}$

$\int \frac{2 \mathrm{dx}}{{x}^{3} \sqrt{{x}^{2} - 1}} = 2 \int \frac{\mathrm{dt}}{\sec} ^ 2 t$

$\int \frac{2 \mathrm{dx}}{{x}^{3} \sqrt{{x}^{2} - 1}} = 2 \int {\cos}^{2} t \mathrm{dt}$

Now:

$2 {\cos}^{2} t = 1 + \cos 2 t$

so:

$\int \frac{2 \mathrm{dx}}{{x}^{3} \sqrt{{x}^{2} - 1}} = \int \left(1 + \cos 2 t\right) \mathrm{dt}$

and using linearity:

$\int \frac{2 \mathrm{dx}}{{x}^{3} \sqrt{{x}^{2} - 1}} = \int \mathrm{dt} + \int \cos 2 t \mathrm{dt}$

$\int \frac{2 \mathrm{dx}}{{x}^{3} \sqrt{{x}^{2} - 1}} = t + \frac{1}{2} \sin 2 t + C$

To undo the substitution note that:

$x = \sec t \implies \tan t = \sqrt{{x}^{2} - 1}$

so:

$t = \arctan \left(\sqrt{{x}^{2} - 1}\right)$

and using the parametric fomulas:

$\sin 2 t = \frac{2 \tan t}{1 + {\tan}^{2} t} = \frac{2 \sqrt{{x}^{2} - 1}}{1 + {x}^{2} - 1} = \frac{2 \sqrt{{x}^{2} - 1}}{x} ^ 2$

So:

$\int \frac{2 \mathrm{dx}}{{x}^{3} \sqrt{{x}^{2} - 1}} = \arctan \left(\sqrt{{x}^{2} - 1}\right) + \frac{\sqrt{{x}^{2} - 1}}{x} ^ 2 + C$

By differentiating we can see that the solution is valid also for $x \in \left(- \infty , - 1\right)$