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

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Lemma for Primitive of $\dfrac 1 {x \left({a x + b}\right)^2}$

$\dfrac 1 {x \left({a x + b}\right)^2} \equiv \dfrac 1 {b^2 x} - \dfrac a {b^2 \left({a x + b}\right)} - \dfrac a {b \left({a x + b}\right)^2}$


Proof

\(\displaystyle \dfrac 1 {x \left({a x + b}\right)^2}\) \(\equiv\) \(\displaystyle \dfrac A x + \dfrac B {a x + b} + \dfrac C {\left({a x + b}\right)^2}\)
\(\text {(1)}: \quad\) \(\displaystyle \implies \ \ \) \(\displaystyle 1\) \(\equiv\) \(\displaystyle A \left({a x + b}\right)^2 + B x \left({a x + b}\right) + C x\) multiplying through by $x \left({a x + b}\right)^2$
\(\displaystyle \) \(\equiv\) \(\displaystyle A a^2 x^2 + 2 A a b x + A b^2 + B a x^2 + B b x + C x\) multiplying everything out


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

\(\displaystyle a x + b\) \(=\) \(\displaystyle 0\)
\(\displaystyle \implies \ \ \) \(\displaystyle x\) \(=\) \(\displaystyle -\frac b a\)
\(\displaystyle \implies \ \ \) \(\displaystyle C \left({-\frac b a}\right)\) \(=\) \(\displaystyle 1\) substituting for $x$ in $(1)$: terms in $a x + b$ are all $0$
\(\displaystyle \implies \ \ \) \(\displaystyle C\) \(=\) \(\displaystyle -\frac a b\)


Equating constants in $(1)$:

\(\displaystyle 1\) \(=\) \(\displaystyle A b^2\)
\(\text {(2)}: \quad\) \(\displaystyle \implies \ \ \) \(\displaystyle A\) \(=\) \(\displaystyle \frac 1 {b^2}\)


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

\(\displaystyle 0\) \(=\) \(\displaystyle A a^2 + B a\)
\(\displaystyle \implies \ \ \) \(\displaystyle B\) \(=\) \(\displaystyle -\frac a {b^2}\) substituting for $A$ from $(2)$ and simplifying


Summarising:

\(\displaystyle A\) \(=\) \(\displaystyle \frac 1 {b^2}\)
\(\displaystyle B\) \(=\) \(\displaystyle -\frac a {b^2}\)
\(\displaystyle C\) \(=\) \(\displaystyle -\frac a b\)

Hence the result.

$\blacksquare$