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

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

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

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

Summarising:

Hence the result.