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

Lemma for Primitive of $\dfrac 1 {x^3 \paren {a x + b} }$

 * $\dfrac 1 {x^3 \paren {a x + b} } \equiv \dfrac {a^2} {b^3 x} + \dfrac {-a} {b^2 x^2} + \dfrac 1 {b x^3} + \dfrac {-a^3} {b^3 \paren {a x + b} }$

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

Equating constants in $(1)$:

Equating $3$rd powers of $x$ in $(1)$:

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

Summarising:

Hence the result.