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

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

 * $\dfrac 1 {x^3 \paren {a^2 - x^2}^2} \equiv \dfrac 1 {a^4 x^3} + \dfrac 2 {a^6 x} + \dfrac {2 x} {a^6 \paren {a^2 - x^2} } + \dfrac x {a^4 \paren {a^2 - x^2}^2}$

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

Setting $x = a$ in $(1)$:

Setting $x = -a$ in $(1)$:

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

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

Equating coefficients of $x^6$ in $(1)$:

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

Summarising:

Thus:

Hence the result.