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

Theorem

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

where $a$ is a constant such that $a > 0$.

Proof
Substitute:


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

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 $\left({-\dfrac \pi 2 .. \dfrac \pi 2}\right)$:


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

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


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

Also see

 * Derivative of Arctangent Function