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

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

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

Proof

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

\(\equiv\)

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

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

\(\ds \leadsto \ \ \)

\(\ds 1\)

\(\equiv\)

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

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

\(\ds \)

\(\equiv\)

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

multiplying everything out

Setting $a x + b = 0$ in $(1)$:

\(\ds a x + b\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds x\)

\(=\)

\(\ds -\frac b a\)

\(\ds \leadsto \ \ \)

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

\(=\)

\(\ds 1\)

substituting for $x$ in $(1)$: terms in $a x + b$ are all $0$

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

\(\ds \leadsto \ \ \)

\(\ds C\)

\(=\)

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

Equating constants in $(1)$:

\(\ds 1\)

\(=\)

\(\ds B b\)

\(\ds \leadsto \ \ \)

\(\ds B\)

\(=\)

\(\ds \frac 1 b\)

Equating $2$nd powers of $x$ in $(1)$:

\(\ds 0\)

\(=\)

\(\ds A a + C\)

\(\ds \leadsto \ \ \)

\(\ds A\)

\(=\)

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

substituting for $C$ from $(2)$ and rearranging

Summarising:

\(\ds A\)

\(=\)

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

\(\ds B\)

\(=\)

\(\ds \frac 1 b\)

\(\ds C\)

\(=\)

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

Hence the result.

$\blacksquare$