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

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

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

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 $a = 1$.

$\blacksquare$