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

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

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

Proof

\(\ds \dfrac 1 {x \paren {a x^2 + b x + c} }\)

\(\equiv\)

\(\ds \frac A x + \frac B x^2 + \frac {C x + D} {a x^2 + b x + c}\)

\(\text {(1)}: \quad\)

\(\ds \leadsto \ \ \)

\(\ds 1\)

\(\equiv\)

\(\ds A \paren {a x^2 + b x + c} x + B \paren {a x^2 + b x + c} + C x^3 + D x^2\)

multiplying through by $x^2 \paren {a x^2 + b x + c}$

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

\(\ds B c\)

\(=\)

\(\ds 1\)

\(\ds \leadsto \ \ \)

\(\ds B\)

\(=\)

\(\ds \frac 1 c\)

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

\(\ds A c + B b\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds A c\)

\(=\)

\(\ds \frac {-b} c\)

\(\ds \leadsto \ \ \)

\(\ds A\)

\(=\)

\(\ds \frac {-b} {c^2}\)

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

\(\ds A a + C\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds C\)

\(=\)

\(\ds \frac {a b} {c^2}\)

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

\(\ds A b + B a + D\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds D\)

\(=\)

\(\ds \frac {b^2 - a c} {c^2}\)

Summarising:

\(\ds A\)

\(=\)

\(\ds \frac {-b} {c^2}\)

\(\ds B\)

\(=\)

\(\ds \frac 1 c\)

\(\ds C\)

\(=\)

\(\ds \frac {a b} {c^2}\)

\(\ds D\)

\(=\)

\(\ds \frac {b^2 - a c} {c^2}\)

Hence the result.

$\blacksquare$