Primitive of Reciprocal of x squared plus a squared/Arctangent Form/Proof 1

Theorem

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

where $a$ is a non-zero constant.

Proof
Let:


 * $a \tan \theta = x$

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

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

Then:

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


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

Thus:


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

When $a = 0$, both $\dfrac x a$ and $\dfrac 1 a$ are undefined.

However, consider the limit of the above primitive as $a \to 0$:

This corresponds with the result:
 * $\displaystyle \int \frac 1 {x^2} \ \mathrm d x = \frac {-1} x + C$

which follows from Primitive of Power.