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

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

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

Proof

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

\(\equiv\)

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

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

\(\ds \leadsto \ \ \)

\(\ds 1\)

\(\equiv\)

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

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

\(\ds \)

\(\equiv\)

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

multiplying everything out

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

\(\ds a x + b\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds x\)

\(=\)

\(\ds -\frac b a\)

\(\ds \leadsto \ \ \)

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

\(=\)

\(\ds 1\)

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

\(\ds \leadsto \ \ \)

\(\ds C\)

\(=\)

\(\ds -\frac a b\)

Equating constants in $(1)$:

\(\ds 1\)

\(=\)

\(\ds A b^2\)

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

\(\ds \leadsto \ \ \)

\(\ds A\)

\(=\)

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

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

\(\ds 0\)

\(=\)

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

\(\ds \leadsto \ \ \)

\(\ds B\)

\(=\)

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

substituting for $A$ from $(2)$ and simplifying

Summarising:

\(\ds A\)

\(=\)

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

\(\ds B\)

\(=\)

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

\(\ds C\)

\(=\)

\(\ds -\frac a b\)

Hence the result.

$\blacksquare$