Primitive of x over a x + b squared

Theorem

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

Proof
Put $u = a x + b$.

Then:

Then: