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

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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.

$\blacksquare$