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

Lemma for Primitive of Reciprocal of $x \paren {x^2 - a^2}$

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

Proof

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

\(\equiv\)

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

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

\(\ds \leadsto \ \ \)

\(\ds 1\)

\(\equiv\)

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

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

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

\(\ds -A a^2\)

\(=\)

\(\ds 1\)

\(\ds \leadsto \ \ \)

\(\ds A\)

\(=\)

\(\ds \frac {-1} {a^2}\)

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

\(\ds 0\)

\(=\)

\(\ds A + B\)

\(\ds \leadsto \ \ \)

\(\ds B\)

\(=\)

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

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

\(\ds 0\)

\(=\)

\(\ds C\)

Summarising:

\(\ds A\)

\(=\)

\(\ds \frac {-1} {a^2}\)

\(\ds B\)

\(=\)

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

\(\ds C\)

\(=\)

\(\ds 0\)

Hence the result.

$\blacksquare$