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

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

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

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

Equating constants in $(1)$:

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

Summarising:

Hence the result.