Primitive of Reciprocal of x by x squared minus a squared squared/Partial Fraction Expansion
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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.
$\blacksquare$