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

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

$\dfrac 1 {x \paren {x^2 + a^2}^2} \equiv \dfrac 1 {a^4 x} - \dfrac x {a^4 \paren {x^2 + a^2} } - \dfrac x {a^2 \paren {x^2 + a^2}^2}$

Proof

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

\(\equiv\)

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

\(\ds \leadsto \ \ \)

\(\ds 1\)

\(\equiv\)

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

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

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

\(\ds \)

\(\equiv\)

\(\ds A x^4 + 2 A a^2 x^2 + A a^4 + B x^4 + B x^2 a^2 + C x^3 + C x a^2 + D x^2 + E x\)

multiplying everything out

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

\(\ds A a^4\)

\(=\)

\(\ds 1\)

\(\ds \leadsto \ \ \)

\(\ds A\)

\(=\)

\(\ds \frac 1 {a^4}\)

Equating coefficients of $x^4$ in $(1)$:

\(\ds 0\)

\(=\)

\(\ds A + B\)

\(\ds \leadsto \ \ \)

\(\ds B\)

\(=\)

\(\ds -\frac 1 {a^4}\)

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

\(\ds C\)

\(=\)

\(\ds 0\)

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

\(\ds 0\)

\(=\)

\(\ds C + E\)

\(\ds \leadsto \ \ \)

\(\ds E\)

\(=\)

\(\ds 0\)

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

\(\ds 0\)

\(=\)

\(\ds 2 A a^2 + B a^2 + D\)

\(\ds \leadsto \ \ \)

\(\ds D\)

\(=\)

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

Summarising:

\(\ds A\)

\(=\)

\(\ds \frac 1 {a^4}\)

\(\ds B\)

\(=\)

\(\ds -\frac 1 {a^4}\)

\(\ds C\)

\(=\)

\(\ds 0\)

\(\ds D\)

\(=\)

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

\(\ds E\)

\(=\)

\(\ds 0\)

Hence the result.

$\blacksquare$