Primitive of Reciprocal of 1 plus x squared/Arctangent Form

Corollary to Primitive of $\frac 1 {x^2 + a^2}$: Arctangent Form

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

where $C$ is an arbitrary constant.

Proof
From Primitive of $\dfrac 1 {x^2 + a^2}$: Arctangent Form:
 * $\ds \int \frac {\d x} {x^2 + a^2} = \frac 1 a \arctan \frac x a + C$

The result follows by setting $a = 1$.