Primitive of Reciprocal of x cubed by x squared minus a squared/Partial Fraction Expansion

Lemma for Primitive of Reciprocal of $x^3 \paren {x^2 - a^2}$

 * $\dfrac 1 {x^3 \paren {x^2 - a^2} } \equiv \dfrac {-1} {a^2 x^3} - \dfrac 1 {a^4 x} + \dfrac x {a^4 \paren {x^2 - a^2} }$

Proof
Setting $x = 0$ in $(1)$:

Equating coefficients of $x$ in $(1)$:

Equating coefficients of $x^2$ in $(1)$:

Equating coefficients of $x^3$ in $(1)$:

Equating coefficients of $x^4$ in $(1)$:

Summarising:

Hence the result.