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

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

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

Proof

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

\(\equiv\)

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

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

\(\ds \leadsto \ \ \)

\(\ds 1\)

\(\equiv\)

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

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

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

\(\ds A c\)

\(=\)

\(\ds 1\)

\(\ds \leadsto \ \ \)

\(\ds A\)

\(=\)

\(\ds \frac 1 c\)

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

\(\ds A b + C\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds C\)

\(=\)

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

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

\(\ds A a + B\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds B\)

\(=\)

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

Summarising:

\(\ds A\)

\(=\)

\(\ds \frac 1 c\)

\(\ds B\)

\(=\)

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

\(\ds C\)

\(=\)

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

Hence the result.

$\blacksquare$