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

Lemma for Primitive of $\dfrac 1 {x \left({a x + b}\right)}$

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

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

Equating constants in $(1)$:

Summarising:

Hence the result.