# \int1/(x^2+a^2)dx?

## "Let $a$ be a constant, and evaluate the integral."

##### 1 Answer
Apr 13, 2018

See process below

#### Explanation:

$\int \frac{1}{{x}^{2} + {a}^{2}} \mathrm{dx}$

Lets do the change $x = a \tan \theta$. With this change we have

$\mathrm{dx} = a {\sec}^{2} \theta d \theta$. Put in the integral

$\int \frac{\cancel{a} {\sec}^{2} \theta d \theta}{{a}^{\cancel{2}} \left({\tan}^{2} \theta + 1\right)}$

But ${\tan}^{2} \theta + 1 = {\sec}^{2} \theta$. The we have

$\int \frac{\cancel{a} {\cancel{\sec}}^{2} \theta d \theta}{{a}^{\cancel{2}} {\cancel{\sec}}^{2} \theta} = \frac{1}{a} \int d \theta = \frac{1}{a} \theta + C$

Undoing the change $\theta = \arctan \left(\frac{x}{a}\right)$. We have finally

$\int \frac{1}{{x}^{2} + {a}^{2}} \mathrm{dx} = \frac{1}{a} \arctan \left(\frac{x}{a}\right) + C$