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

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

\(\equiv\)

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

\(\ds \leadsto \ \ \)

\(\ds 1\)

\(\equiv\)

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

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

\(\ds \)

\(\)

\(\, \ds + \, \)

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

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

\(\ds \leadsto \ \ \)

\(\ds 1\)

\(\equiv\)

\(\ds A a^2 x^4 + 2 A a b x^3 + 2 A a c x^2\)

multiplying out

\(\ds \)

\(\)

\(\, \ds + \, \)

\(\ds A b^2 x^2 + 2 A b c x + A c^2\)

\(\ds \)

\(\)

\(\, \ds + \, \)

\(\ds B a x^4 + B b x^3 + B c x^2\)

\(\ds \)

\(\)

\(\, \ds + \, \)

\(\ds C a x^3 + C b x^2 + C c x\)

\(\ds \)

\(\)

\(\, \ds + \, \)

\(\ds D x^2 + E x\)

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

\(\ds A c^2\)

\(=\)

\(\ds 1\)

\(\text {(2)}: \quad\)

\(\ds \leadsto \ \ \)

\(\ds A\)

\(=\)

\(\ds \frac 1 {c^2}\)

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

\(\ds A a^2 + B a\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

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

\(=\)

\(\ds -B a\)

substituting for $A$ from $(2)$

\(\text {(3)}: \quad\)

\(\ds \leadsto \ \ \)

\(\ds B\)

\(=\)

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

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

\(\ds 2 A a b + B b + C a\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds 2 \frac 1 {c^2} a b + \frac {-a} {c^2} b\)

\(=\)

\(\ds -C a\)

substituting for $A$ from $(2)$ and $B$ from $(3)$

\(\ds \leadsto \ \ \)

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

\(=\)

\(\ds -C\)

simplifying

\(\text {(4)}: \quad\)

\(\ds \leadsto \ \ \)

\(\ds C\)

\(=\)

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

simplifying

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

\(\ds 2 A b c + C c + E\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds 2 \frac 1 {c^2} b c + \frac {-b} {c^2} c\)

\(=\)

\(\ds -E\)

\(\ds \leadsto \ \ \)

\(\ds E\)

\(=\)

\(\ds -2 \frac b c + \frac b c\)

\(\ds \leadsto \ \ \)

\(\ds E\)

\(=\)

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

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

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

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds \frac 1 {c^2} \paren {2 a c + b^2} + \frac {-a} {c^2} c + \frac {-b} {c^2} b\)

\(=\)

\(\ds - D\)

\(\ds \leadsto \ \ \)

\(\ds D\)

\(=\)

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

\(\ds \leadsto \ \ \)

\(\ds D\)

\(=\)

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

Summarising:

\(\ds A\)

\(=\)

\(\ds \frac 1 {c^2}\)

\(\ds B\)

\(=\)

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

\(\ds C\)

\(=\)

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

\(\ds D\)

\(=\)

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

\(\ds E\)

\(=\)

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

Hence the result.

$\blacksquare$