# How do you evaluate int dx/(x^2+4x+13) from [-2, 2]?

Nov 24, 2016

${\int}_{- 2}^{2} \frac{\mathrm{dx}}{{x}^{2} + 4 x + 13} = \frac{1}{3} \arctan \left(\frac{4}{3}\right)$

#### Explanation:

First without the bounds:

$I = \int \frac{\mathrm{dx}}{{x}^{2} + 4 x + 13}$

Complete the square in the denominator.

$I = \int \frac{\mathrm{dx}}{{\left(x + 2\right)}^{2} + 9}$

Now we can use trigonometric substitution. Let $x + 2 = 3 \tan \theta$. This also implies that $\mathrm{dx} = 3 {\sec}^{2} \theta d \theta$.

$I = \int \frac{3 {\sec}^{2} \theta d \theta}{9 {\tan}^{2} \theta + 9} = \frac{1}{3} \int \frac{{\sec}^{2} \theta d \theta}{{\tan}^{2} \theta + 1} = \frac{1}{3} \int \frac{{\sec}^{2} \theta d \theta}{\sec} ^ 2 \theta$

$I = \frac{1}{3} \int d \theta = \frac{1}{3} \theta$

From $x + 2 = 3 \tan \theta$ we see that $\theta = \arctan \left(\frac{x + 2}{3}\right)$.

$I = \frac{1}{3} \arctan \left(\frac{x + 2}{3}\right)$

So now applying the bounds:

${I}_{B} = {\int}_{- 2}^{2} \frac{\mathrm{dx}}{{x}^{2} + 4 x + 13} = {\left[\frac{1}{3} \arctan \left(\frac{x + 2}{3}\right)\right]}_{- 2}^{2}$

So

${I}_{B} = \left(\frac{1}{3} \arctan \left(\frac{2 + 2}{3}\right)\right) - \left(\frac{1}{3} \arctan \left(\frac{- 2 + 2}{3}\right)\right)$

${I}_{B} = \frac{1}{3} \arctan \left(\frac{4}{3}\right)$