# Integrate the following int sqrt(x^2+81) dx?

Feb 7, 2017

$\frac{x \sqrt{{x}^{2} + 81}}{2} + \frac{81}{2} \ln | \frac{x + \sqrt{{x}^{2} + 81}}{9} | + C$

#### Explanation:

Use trig substitution. Let $x = 9 \tan \theta$. Then $\mathrm{dx} = 9 {\sec}^{2} \theta d \theta$.

$\implies \int \sqrt{{\left(9 \tan \theta\right)}^{2} + 81} \cdot 9 {\sec}^{2} \theta d \theta$

$\implies \int \sqrt{81 {\tan}^{2} \theta + 81} \cdot 9 {\sec}^{2} \theta d \theta$

$\implies \int \sqrt{81 \left({\tan}^{2} \theta + 1\right)} \cdot 9 {\sec}^{2} \theta d \theta$

$\implies \int \sqrt{81 {\sec}^{2} \theta} \cdot 9 {\sec}^{2} \theta d \theta$

$\implies \int 9 \sec \theta \cdot 9 {\sec}^{2} \theta d \theta$

$\implies \int 81 {\sec}^{3} \theta d \theta$

$\implies 81 \int {\sec}^{3} \theta d \theta$

This is a relatively known integral. The proof can be found here.

$\implies \frac{81}{2} \sec \theta \tan \theta + \frac{81}{2} \ln | \sec \theta + \tan \theta | + C$

We know from our initial substitution that $\tan \theta = \frac{x}{9}$. This means the hypotenuse of the triangle would have a hypotenuse of $\sqrt{{x}^{2} + 81}$.

$\implies \frac{81}{2} \left(\frac{\sqrt{{x}^{2} + 81}}{9}\right) \left(\frac{x}{9}\right) + \frac{81}{2} \ln | \frac{\sqrt{{x}^{2} + 81}}{9} + \frac{x}{9} | + C$

$\implies \frac{x \sqrt{{x}^{2} + 81}}{2} + \frac{81}{2} \ln | \frac{x + \sqrt{{x}^{2} + 81}}{9} | + C$

Hopefully this helps!