Primitive of Reciprocal of x squared plus a squared/Arctangent Form/Proof 2

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
We have that $x^2 + a^2$ is in the form $a x^2 + b x + c$, where $b^2 - 4 a c < 0$.

Thus from Primitive of $\dfrac 1 {a x^2 + b x + c}$ for $b^2 - 4 a c > 0$:
 * $\displaystyle \int \frac {\mathrm d x} {a x^2 + b x + c} = \frac 2 {\sqrt {4 a c - b^2} } \arctan \left({\frac {2 a x + b} {\sqrt {4 a c - b^2} } }\right) + C$

setting $a := 1, b := 0, c := a^2$: