Primitive of Reciprocal of a x + b squared by p x + q/Partial Fraction Expansion

Lemma for Primitive of Reciprocal of a x + b squared by p x + q

 * $\dfrac 1 {\left({a x + b}\right)^2 \left({p x + q}\right)} \equiv \dfrac 1 {b p - a q} \left({\dfrac {-a p} {\left({b p - a q}\right) \left({a x + b}\right)} + \dfrac {-a} {\left({a x + b}\right)^2} + \dfrac {p^2} {\left({b p - a q}\right) \left({p x + q}\right)} }\right)$

Proof
Setting $a x + b = 0$ in $(1)$:

Setting $p x + q = 0$ in $(1)$:

Equating $2$nd powers of $x$ in $(1)$:

Summarising:

Hence the result.