# How do you integrate int ( 1/((x+1)^2+4))  using partial fractions?

Sep 12, 2016

$\frac{1}{2} \arctan \left(\frac{x + 1}{2}\right) + C$

#### Explanation:

This cannot be expressed using partial fractions. However, we can integrate this using trigonometric substitutions.

$\int \frac{\mathrm{dx}}{{\left(x + 1\right)}^{2} + 4}$

Let $x + 1 = 2 \tan \theta$. Thus, $\mathrm{dx} = 2 {\sec}^{2} \theta d \theta$. Substituting, this gives us:

$= \int \frac{2 {\sec}^{2} \theta d \theta}{4 {\tan}^{2} \theta + 4}$

Factoring:

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

Recall that ${\tan}^{2} \theta + 1 = {\sec}^{2} \theta$:

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

$= \frac{1}{2} \int d \theta$

$= \frac{1}{2} \theta + C$

From $x + 1 = 2 \tan \theta$ we see that $\theta = \arctan \left(\frac{x + 1}{2}\right)$:

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