# How do you integrate sqrt(x^2-9)/(x)dx?

Feb 17, 2015

To integrate this function we will use a trigonometric substitution.

Let $\cos \theta = \frac{3}{x}$

Therefore, $\sec \theta = \frac{x}{3}$ and $x = 3 \sec \theta$

Differentiate $x = 3 \sec \theta$

$\mathrm{dx} = 3 \sec \theta \tan \theta d \theta$

Make the substitution into the integral

$\int \frac{\sqrt{{\left(3 \sec \theta\right)}^{2} - 9}}{3 \sec \theta} 3 \sec \theta \tan \theta d \theta$

$\int \sqrt{9 {\sec}^{2} \theta - 9} \tan \theta d \theta$

$\int \sqrt{9 \left({\sec}^{2} \theta - 1\right)} \tan \theta d \theta$

$\int \sqrt{9} \sqrt{{\tan}^{2} \theta} \tan \theta d \theta$

$3 \int {\tan}^{2} \theta d \theta$

$3 \int {\sec}^{2} \theta - 1 d \theta$

Now integrating we get

$3 \left(\tan \theta - \theta\right)$

Now define $\theta$ and $\tan \theta$ in terms of $x$ as follows

$\theta = \arctan \left(\frac{\sqrt{{x}^{2} - 9}}{3}\right)$

$\tan \theta = \frac{\sqrt{{x}^{2} - 9}}{3}$

now back substitute

$3 \left(\frac{\sqrt{{x}^{2} - 9}}{3} - \arctan \left(\frac{\sqrt{{x}^{2} - 9}}{3}\right)\right) + C$

Distributing the $3$ we will have

$\sqrt{{x}^{2} - 9} - 3 \arctan \left(\frac{\sqrt{{x}^{2} - 9}}{3}\right) + C$ FINAL ANSWER