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

## 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.

$\blacksquare$