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

## Lemma for Primitive of Reciprocal of $x \left({x^2 - a^2}\right)$

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

## Proof

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

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

 $\displaystyle -A a^2$ $=$ $\displaystyle 1$ $\displaystyle \implies \ \$ $\displaystyle A$ $=$ $\displaystyle \frac {-1} {a^2}$

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

 $\displaystyle 0$ $=$ $\displaystyle A + B$ $\displaystyle \implies \ \$ $\displaystyle B$ $=$ $\displaystyle \frac 1 {a^2}$

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

 $\displaystyle 0$ $=$ $\displaystyle C$

Summarising:

 $\displaystyle A$ $=$ $\displaystyle \frac {-1} {a^2}$ $\displaystyle B$ $=$ $\displaystyle \frac 1 {a^2}$ $\displaystyle C$ $=$ $\displaystyle 0$

Hence the result.

$\blacksquare$