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

Lemma for Primitive of Reciprocal of $x^3 \paren {x^2 + a^2}$

$\dfrac 1 {x^3 \paren {x^2 + a^2} } \equiv \dfrac 1 {a^2 x^3} - \dfrac 1 {a^4 x} + \dfrac x {a^4 \paren {x^2 + a^2} }$

Proof

\(\ds \dfrac 1 {x^3 \paren {x^2 + a^2} }\)

\(\equiv\)

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

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

\(\ds \leadsto \ \ \)

\(\ds 1\)

\(\equiv\)

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

multiplying through by $x^3 \paren {x^2 + a^2}$

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

\(\ds C a^2\)

\(=\)

\(\ds 1\)

\(\ds \leadsto \ \ \)

\(\ds C\)

\(=\)

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

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

\(\ds 0\)

\(=\)

\(\ds B a^2\)

\(\ds \leadsto \ \ \)

\(\ds B\)

\(=\)

\(\ds 0\)

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

\(\ds 0\)

\(=\)

\(\ds A a^2 + C\)

\(\ds \leadsto \ \ \)

\(\ds A\)

\(=\)

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

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

\(\ds 0\)

\(=\)

\(\ds B + E\)

\(\ds \leadsto \ \ \)

\(\ds E\)

\(=\)

\(\ds 0\)

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

\(\ds 0\)

\(=\)

\(\ds A + D\)

\(\ds \leadsto \ \ \)

\(\ds D\)

\(=\)

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

Summarising:

\(\ds A\)

\(=\)

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

\(\ds B\)

\(=\)

\(\ds 0\)

\(\ds C\)

\(=\)

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

\(\ds D\)

\(=\)

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

\(\ds E\)

\(=\)

\(\ds 0\)

Hence the result.

$\blacksquare$