# Question #fcb11

Aug 31, 2017

You are correct: $\int \frac{\sqrt{{x}^{2} - 1}}{x} ^ 2 \mathrm{dx} = \ln \left\mid x + \sqrt{{x}^{2} - 1} \right\mid - \frac{\sqrt{{x}^{2} - 1}}{x} + C$

#### Explanation:

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

I'm sure there are other methods of approach, but when I see square roots I immediately think of trig substitutions. Let's try $x = \sec \theta$, implying that ${x}^{2} - 1 = {\sec}^{2} \theta - 1 = {\tan}^{2} \theta$ and that $\mathrm{dx} = \sec \theta \tan \theta d \theta$:

$I = \int \frac{\sqrt{{\tan}^{2} \theta}}{\sec} ^ 2 \theta \left(\sec \theta \tan \theta d \theta\right)$

$I = \int {\tan}^{2} \frac{\theta}{\sec} \theta d \theta$

$I = \int \frac{{\sec}^{2} \theta - 1}{\sec} \theta d \theta$

$I = \int \left(\sec \theta - \cos \theta\right) d \theta$

$I = \ln \left\mid \sec \theta + \tan \theta \right\mid - \sin \theta + C$

The substitution $x = \sec \theta$ implies the following:

• $\tan \theta = \sqrt{{\sec}^{2} \theta - 1} = \sqrt{{x}^{2} - 1}$
• $\cos \theta = \frac{1}{x}$
• $\sin \theta = \sqrt{1 - {\cos}^{2} \theta} = \sqrt{1 - \frac{1}{x} ^ 2} = \frac{\sqrt{{x}^{2} - 1}}{x}$

Thus:

$I = \ln \left\mid x + \sqrt{{x}^{2} - 1} \right\mid - \frac{\sqrt{{x}^{2} - 1}}{x} + C$

You were right!