Primitive of Reciprocal of x squared by a squared minus x squared squared/Partial Fraction Expansion
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Lemma for Primitive of Reciprocal of $x^2 \paren {a^2 - x^2}^2$
- $\dfrac 1 {x^2 \paren {a^2 - x^2}^2} \equiv \dfrac 1 {a^4 x^2} + \dfrac 3 {4 a^5 \paren {a + x} } - \dfrac 3 {4 a^5 \paren {a - x} } + \dfrac 1 {4 a^4 \paren {a + x}^2} + \dfrac 1 {4 a^4 \paren {a - x}^2}$
Proof
\(\ds \frac 1 {x^2 \paren {a^2 - x^2}^2}\) | \(=\) | \(\ds \frac 1 {x^2 \paren {a + x}^2 \paren {a - x}^2}\) | Difference of Two Squares | |||||||||||
\(\ds \) | \(\equiv\) | \(\ds \frac A {a + x} + \frac B {\paren {a + x}^2} + \frac C {a - x} + \frac D {\paren {a - x}^2} + \frac E x + \frac F {x^2}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds 1\) | \(\equiv\) | \(\ds A x^2 \paren {a^2 - x^2} \paren {a - x} + B x^2 \paren {a - x}^2\) | multiplying through by $x^2 \paren {a^2 - x^2}^2$ | ||||||||||
\(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds C x^2 \paren {a^2 - x^2} \paren {a + x} + D x^2 \paren {a + x}^2\) | |||||||||||
\(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds E x \paren {a^2 - x^2}^2 + F \paren {a^2 - x^2}^2\) | |||||||||||
\(\text {(1)}: \quad\) | \(\ds \) | \(\equiv\) | \(\ds A x^5 - A a x^4 - A a^2 x^3 + A a^3 x^2 + B x^4 - 2 B a x^3 + B a^2 x^2\) | multiplying out | ||||||||||
\(\ds \) | \(\) | \(\, \ds - \, \) | \(\ds C x^5 - C a x^4 + C a^2 x^3 + C a^3 x^2 + D x^4 + 2 D a x^3 + D a^2 x^2\) | |||||||||||
\(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds E x^5 - 2 E a^2 x^3 + E a^4 x + F x^4 - 2 F a^2 x^2 + F a^4\) |
Setting $x = a$ in $(1)$:
\(\ds D a^2 \paren {2 a}^2\) | \(=\) | \(\ds 1\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds D\) | \(=\) | \(\ds \frac 1 {4 a^4}\) |
Setting $x = -a$ in $(1)$:
\(\ds B \paren {-a}^2 \paren {2 a}^2\) | \(=\) | \(\ds 1\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds B\) | \(=\) | \(\ds \frac 1 {4 a^4}\) |
Setting $x = 0$ in $(1)$:
\(\ds F a^4\) | \(=\) | \(\ds 1\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds F\) | \(=\) | \(\ds \frac 1 {a^4}\) |
Equating coefficients of $x$ in $(1)$:
\(\ds 0\) | \(=\) | \(\ds E\) |
Equating coefficients of $x^5$ in $(1)$:
\(\text {(2)}: \quad\) | \(\ds 0\) | \(=\) | \(\ds A - C + E\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds 0\) | \(=\) | \(\ds A - C\) | as $E = 0$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds A\) | \(=\) | \(\ds C\) |
Equating coefficients of $x^2$ in $(1)$:
\(\ds A a^3 + B a^2 + C a^3 + D a^2 - 2 F a^2\) | \(=\) | \(\ds 0\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds A a^3 + \frac 1 {4 a^4} a^2 + C a^3 + \frac 1 {4 a^4} a^2 - 2 \frac 1 {a^4} a^2\) | \(=\) | \(\ds 0\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds A + \frac 1 {4 a^5} + C + \frac 1 {4 a^5} - \frac 8 {4 a^5}\) | \(=\) | \(\ds 0\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds A + C\) | \(=\) | \(\ds \frac {8 - 1 - 1} {4 a^5}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 3 {2 a^5}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds A\) | \(=\) | \(\ds \frac 3 {4 a^5}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds C\) | \(=\) | \(\ds \frac 3 {4 a^5}\) | as $A = C$ |
Summarising:
\(\ds A\) | \(=\) | \(\ds \frac 3 {4 a^5}\) | ||||||||||||
\(\ds B\) | \(=\) | \(\ds \frac 1 {4 a^4}\) | ||||||||||||
\(\ds C\) | \(=\) | \(\ds \frac 3 {4 a^5}\) | ||||||||||||
\(\ds D\) | \(=\) | \(\ds \frac 1 {4 a^4}\) | ||||||||||||
\(\ds E\) | \(=\) | \(\ds 0\) | ||||||||||||
\(\ds F\) | \(=\) | \(\ds \frac 1 {a^4}\) |
Hence the result.
$\blacksquare$