Primitive of Reciprocal of a x + b squared/Proof 2

Theorem

 * $\displaystyle \int \frac {\mathrm d x} {\left({a x + b}\right)^2} = -\frac 1 {a \left({a x + b}\right)} + C$

Proof
From Primitive of Power of $a x + b$:
 * $\displaystyle \int \left({a x + b}\right)^n \ \mathrm d x = \frac {\left({a x + b}\right)^{n + 1} } {\left({n + 1}\right) a} + C$

where $n \ne 1$.

The result follows by setting $n = -2$.