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

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

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

Proof

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

\(\equiv\)

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

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

\(\ds \leadsto \ \ \)

\(\ds 1\)

\(\equiv\)

\(\ds A x \paren {x^2 - a^2} + B \paren {x^2 - a^2} + C x^3 + D x^2\)

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

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

\(\ds -B a^2\)

\(=\)

\(\ds 1\)

\(\ds \leadsto \ \ \)

\(\ds B\)

\(=\)

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

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

\(\ds 0\)

\(=\)

\(\ds -A a^2\)

\(\ds \leadsto \ \ \)

\(\ds A\)

\(=\)

\(\ds 0\)

Equating coefficients of $x^3$ in $(1)$:

\(\ds 0\)

\(=\)

\(\ds A + C\)

\(\ds \leadsto \ \ \)

\(\ds C\)

\(=\)

\(\ds 0\)

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

\(\ds 0\)

\(=\)

\(\ds B + D\)

\(\ds \leadsto \ \ \)

\(\ds D\)

\(=\)

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

Summarising:

\(\ds A\)

\(=\)

\(\ds 0\)

\(\ds B\)

\(=\)

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

\(\ds C\)

\(=\)

\(\ds 0\)

\(\ds D\)

\(=\)

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

Hence the result.

$\blacksquare$