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

1 Answer
Nov 9, 2016

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

Explanation:

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

This isn't a job for partial fractions, it's a job for a trig substitution. Let $x + 1 = 2 \tan \theta$. This implies that $\mathrm{dx} = 2 {\sec}^{2} \theta d \theta$. Substituting in tells us:

$I = \int \frac{1}{4 {\tan}^{2} \theta + 4} \left(2 {\sec}^{2} \theta d \theta\right) = \frac{1}{2} \int {\sec}^{2} \frac{\theta}{1 + {\tan}^{2} \theta} d \theta$

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

$I = \frac{1}{2} \int d \theta = \frac{1}{2} \theta + C$

Undoing the substitution $x + 1 = 2 \tan \theta$:

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