Primitive of Reciprocal of x by square of a x squared plus b x plus c/Partial Fraction Expansion

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

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

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

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

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

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

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

Summarising:

Hence the result.