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

Corollary to Primitive of Reciprocal of x squared plus a squared

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

where $a$ and $b$ are non-zero constants.

Proof
Let $z = b x$.

Then:
 * $\dfrac {\d x} {\d z} = \dfrac 1 b$

Hence: