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

Lemma for Primitive of $\dfrac 1 {x \paren {a x + b}^2}$

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

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

Equating constants in $(1)$:

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

Summarising:

Hence the result.