Primitive of Reciprocal of x by x squared minus 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

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

\(=\)

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

Difference of Two Squares

\(\ds \)

\(\equiv\)

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

\(\ds \leadsto \ \ \)

\(\ds 1\)

\(\equiv\)

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

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

\(\ds \)

\(\)

\(\, \ds + \, \)

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

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

\(\ds \)

\(\equiv\)

\(\ds A x^4 - A a x^3 - A a^2 x^2 + A a^3 x + B x^3 - 2 B a x^2 + B a^2 x\)

multiplying out

\(\ds \)

\(\)

\(\, \ds + \, \)

\(\ds C x^4 + C a x^3 - C a^2 x^2 - C a^3 x + D x^3 + 2 D a x^2 + D a^2 x\)

\(\ds \)

\(\)

\(\, \ds + \, \)

\(\ds E x^4 - 2 E a^2 x^2 + E a^4\)

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

\(\ds D a \paren {2 a}^2\)

\(=\)

\(\ds 1\)

\(\ds \leadsto \ \ \)

\(\ds D\)

\(=\)

\(\ds \frac 1 {4 a^3}\)

Setting $x = -a$ in $(1)$:

\(\ds B \paren {-a} \paren {-2 a}^2\)

\(=\)

\(\ds 1\)

\(\ds \leadsto \ \ \)

\(\ds B\)

\(=\)

\(\ds \frac {-1} {4 a^3}\)

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

\(\ds E a^4\)

\(=\)

\(\ds 1\)

\(\ds \leadsto \ \ \)

\(\ds E\)

\(=\)

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

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

\(\ds 0\)

\(=\)

\(\ds A + C + E\)

\(\ds \leadsto \ \ \)

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

\(=\)

\(\ds C\)

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

\(\ds - A a + C a + B + D\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds - A a + C a + \frac 1 {4 a^3} + \frac {-1} {4 a^3}\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

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

\(=\)

\(\ds 0\)

as $C = -A -\dfrac 1 {a^4}$

\(\ds \leadsto \ \ \)

\(\ds -2 A - \frac 1 {a^4}\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds A\)

\(=\)

\(\ds \frac {-1} {2 a^4}\)

\(\ds \leadsto \ \ \)

\(\ds C\)

\(=\)

\(\ds \frac {-1} {2 a^4}\)

Summarising:

\(\ds A\)

\(=\)

\(\ds \frac {-1} {2 a^4}\)

\(\ds B\)

\(=\)

\(\ds \frac {-1} {4 a^3}\)

\(\ds C\)

\(=\)

\(\ds \frac {-1} {2 a^4}\)

\(\ds D\)

\(=\)

\(\ds \frac 1 {4 a^3}\)

\(\ds E\)

\(=\)

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

Thus:

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

\(\equiv\)

\(\ds \frac {-1} {2 a^4 \paren {x + a} } + \frac {-1} {4 a^3 \paren {x + a}^2} + \frac {-1} {2 a^4 \paren {x - a} } + \frac 1 {4 a^3 \paren {x - a}^2} + \frac 1 {a^4 x}\)

\(\ds \)

\(\equiv\)

\(\ds - \frac {\paren {x - a} + \paren {x + a} } {2 a^4 \paren {x + a} \paren {x - a} } + \frac {\paren {x + a}^2 - \paren {x - a}^2} {4 a^3 \paren {x + a}^2 \paren {x - a}^2} + \frac 1 {a^4 x}\)

common denominators

\(\ds \)

\(\equiv\)

\(\ds \frac {- 2 x} {2 a^4 \paren {x^2 - a^2} } + \frac {\paren {x^2 + 2 a x + a^2} - \paren {x^2 - 2 a x + a^2} } {4 a^3 \paren {x^2 - a^2}^2} + \frac 1 {a^4 x}\)

simplifying

\(\ds \)

\(\equiv\)

\(\ds \frac {-x} {a^4 \paren {x^2 - a^2} } + \frac {4 a x} {4 a^3 \paren {x^2 - a^2}^2} + \frac 1 {a^4 x}\)

simplifying

\(\ds \)

\(\equiv\)

\(\ds \dfrac 1 {a^4 x} + \dfrac {-x} {a^4 \paren {x^2 - a^2} } + \dfrac x {a^2 \paren {x^2 - a^2}^2}\)

simplifying and rearranging

Hence the result.

$\blacksquare$