Primitive of Reciprocal of x squared plus a squared/Arctangent Form

Theorem

 * $\displaystyle \int \frac {1}{x^2 + a^2} \ \mathrm dx = \frac {1}{a}\arctan{\frac{x}{a}} + C$

where $a$ is a constant and $a > 0$.

Proof
Substitute:


 * $\tan \theta = \dfrac {x}{a} \iff x = a \tan \theta$

for $\theta \in \left(-\dfrac {\pi}{2}..\dfrac {\pi}{2}\right)$.

From Shape of Tangent Function, this substitution is valid for all real $x$.

As $\theta$ was stipulated to be in the open interval $\left(-\dfrac {\pi}{2}..\dfrac {\pi}{2}\right)$:


 * $\tan \theta = \dfrac {x}{a} \iff \theta = \arctan \dfrac{x}{a}$

The answer in terms of $x$, then, is:


 * $\displaystyle \int \frac {1}{x^2 + a^2} \ \mathrm dx = \frac {1}{a}\arctan{\frac{x}{a}} + C$