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

 * $\dfrac 1 {\paren {a x + b}^2 \paren {p x + q} } \equiv \dfrac 1 {b p - a q} \paren {\dfrac {-a p} {\paren {b p - a q} \paren {a x + b} } + \dfrac {-a} {\paren {a x + b}^2} + \dfrac {p^2} {\paren {b p - a q} \paren {p x + q} } }$

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.