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

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

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

for $b p \ne a q$.

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

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

Summarising:

Hence the result.