# How do you evaluate the integral int dx/(4+x^2)?

##### 1 Answer
Jan 15, 2017

You may know the following rule:

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

So:

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

We can derive the $\int \frac{\mathrm{dx}}{{a}^{2} + {x}^{2}}$ rule. To do so, let $x = a \tan \theta$.

Thus, $\mathrm{dx} = a {\sec}^{2} \theta d \theta$ and ${x}^{2} = {a}^{2} {\tan}^{2} \theta$.

$\int \frac{\mathrm{dx}}{{a}^{2} + {x}^{2}} = \int \frac{a {\sec}^{2} \theta d \theta}{{a}^{2} + {a}^{2} {\tan}^{2} \theta}$

Factoring the ${a}^{2}$ terms and using the identity ${\tan}^{2} \theta + 1 = {\sec}^{2} \theta$:

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

The antiderivative of $1$ is $\theta$, since we're working in terms of $\theta$ here:

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

From $x = a \tan \theta$, our original substitution, we can solve for $\theta$. Note that $\tan \theta = \frac{x}{a}$ so $\theta = \arctan \left(\frac{x}{a}\right)$. Then:

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

Which is the above stated rule.