Primitive of Reciprocal of x squared minus a squared squared/Partial Fraction Expansion
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Lemma for Primitive of Reciprocal of $\paren {x^2 - a^2}^2$
- $\dfrac 1 {\paren {x^2 - a^2}^2} \equiv \dfrac 1 {4 a^3 \paren {x + a} } - \dfrac 1 {4 a^3 \paren {x - a} } + \dfrac 1 {4 a^2 \paren {x + a}^2} + \dfrac 1 {4 a^2 \paren {x - a}^2}$
Proof
| \(\ds \dfrac 1 {\paren {x^2 - a^2}^2}\) | \(=\) | \(\ds \dfrac 1 {\paren {x + a}^2 \paren {x - a}^2}\) | Difference of Two Squares | |||||||||||
| \(\ds \) | \(\equiv\) | \(\ds \dfrac A {x + a} + \dfrac B {\paren {x + a}^2} + \dfrac C {x - a} + \dfrac D {\paren {x - a}^2}\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 1\) | \(\equiv\) | \(\ds A \paren {x^2 - a^2} \paren {x - a} + B \paren {x - a}^2 + C \paren {x^2 - a^2} \paren {x + a} + D \paren {x + a}^2\) | multiplying through by $\paren {x^2 - a^2}^2$ | ||||||||||
| \(\text {(1)}: \quad\) | \(\ds \) | \(\equiv\) | \(\ds A x^3 - A a x^2 - A a^2 x + A a^3 + B x^2 - 2 B a x + B a^2\) | multiplying out | ||||||||||
| \(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds C x^3 + C a x^2 - C a^2 x - C a^3 + D x^2 + 2 D a x + D a^2\) |
Setting $x = a$ in $(1)$:
| \(\ds D \paren {2 a}^2\) | \(=\) | \(\ds 1\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds D\) | \(=\) | \(\ds \frac 1 {4 a^2}\) |
Setting $x = -a$ in $(1)$:
| \(\ds B \paren {-2 a}^2\) | \(=\) | \(\ds 1\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds B\) | \(=\) | \(\ds \frac 1 {4 a^2}\) |
Equating coefficients of $x^3$ in $(1)$:
| \(\ds 0\) | \(=\) | \(\ds A + C\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds A\) | \(=\) | \(\ds -C\) |
Equating coefficients of $x^2$ in $(1)$:
| \(\ds - A a + C a + B + D\) | \(=\) | \(\ds 0\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds - A a + C a + \frac 1 {4 a^2} + \frac 1 {4 a^2}\) | \(=\) | \(\ds 0\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds C a + C a + \frac 1 {2 a^2}\) | \(=\) | \(\ds 0\) | as $A = - C$ | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 2 C\) | \(=\) | \(\ds \frac {-1} {2 a^3}\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds C\) | \(=\) | \(\ds \frac {-1} {4 a^3}\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds A\) | \(=\) | \(\ds \frac 1 {4 a^3}\) |
Summarising:
| \(\ds A\) | \(=\) | \(\ds \frac 1 {4 a^3}\) | ||||||||||||
| \(\ds B\) | \(=\) | \(\ds \frac 1 {4 a^2}\) | ||||||||||||
| \(\ds C\) | \(=\) | \(\ds \frac {-1} {4 a^3}\) | ||||||||||||
| \(\ds D\) | \(=\) | \(\ds \frac 1 {4 a^2}\) |
Hence the result.
$\blacksquare$