Primitive of x over a x + b squared by p x + q/Corollary

Theorem

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