Primitive of Reciprocal of x cubed by x squared minus a squared squared/Partial Fraction Expansion
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Lemma for Primitive of Reciprocal of $x^3 \paren {x^2 - a^2}^2$
- $\dfrac 1 {x^3 \paren {x^2 - a^2}^2} \equiv \dfrac 1 {a^4 x^3} + \dfrac 2 {a^6 x} - \dfrac {2 x} {a^6 \paren {x^2 - a^2} } + \dfrac x {a^4 \paren {x^2 - a^2}^2}$
Proof
| \(\ds \frac 1 {x^3 \paren {x^2 - a^2}^2}\) | \(=\) | \(\ds \frac 1 {x^3 \paren {x + a}^2 \paren {x - a}^2}\) | Difference of Two Squares | |||||||||||
| \(\ds \) | \(\equiv\) | \(\ds \frac A {x + a} + \frac B {\paren {x + a}^2} + \frac C {x - a} + \frac D {\paren {x - a}^2} + \frac E x + \frac F {x^2} + \frac G {x^3}\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 1\) | \(\equiv\) | \(\ds A x^3 \paren {x^2 - a^2} \paren {x - a} + B x^3 \paren {x - a}^2\) | multiplying through by $x^3 \paren {x^2 - a^2}^2$ | ||||||||||
| \(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds C x^3 \paren {x^2 - a^2} \paren {x + a} + D x^3 \paren {x + a}^2\) | |||||||||||
| \(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds E x^2 \paren {x^2 - a^2}^2 + F x \paren {x^2 - a^2}^2 + G \paren {x^2 - a^2}^2\) | |||||||||||
| \(\text {(1)}: \quad\) | \(\ds \) | \(\equiv\) | \(\ds A x^6 - A a x^5 - A a^2 x^4 + A a^3 x^3 + B x^5 - 2 B a x^4 + B a^2 x^3\) | multiplying out | ||||||||||
| \(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds C x^6 + C a x^5 - C a^2 x^4 - C a^3 x^3 + D x^5 + 2 D a x^4 + D a^2 x^3\) | |||||||||||
| \(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds E x^6 - 2 E a^2 x^4 + E a^4 x^2 + F x^5 - 2 F a^2 x^3 + F x a^4\) | |||||||||||
| \(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds G x^4 - 2 G a^2 x^2 + G a^4\) |
Setting $x = 0$ in $(1)$:
| \(\ds G a^4\) | \(=\) | \(\ds 1\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds G\) | \(=\) | \(\ds \frac 1 {a^4}\) |
Setting $x = a$ in $(1)$:
| \(\ds D a^3 \paren {2 a}^2\) | \(=\) | \(\ds 1\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds D\) | \(=\) | \(\ds \frac 1 {4 a^5}\) |
Setting $x = -a$ in $(1)$:
| \(\ds B \paren {-a}^3 \paren {-2 a}^2\) | \(=\) | \(\ds 1\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds B\) | \(=\) | \(\ds \frac {-1} {4 a^5}\) |
Equating coefficients of $x$ in $(1)$:
| \(\ds 0\) | \(=\) | \(\ds F\) |
Equating coefficients of $x^2$ in $(1)$:
| \(\ds 0\) | \(=\) | \(\ds E a^4 - 2 G a^2\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds E a^4\) | \(=\) | \(\ds 2 \frac 1 {a^4} a^2\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds E\) | \(=\) | \(\ds \frac 2 {a^6}\) |
Equating coefficients of $x^6$ in $(1)$:
| \(\text {(2)}: \quad\) | \(\ds 0\) | \(=\) | \(\ds A + C + E\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 0\) | \(=\) | \(\ds A + C + \frac 2 {a^6}\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds A + C\) | \(=\) | \(\ds -\frac 2 {a^6}\) |
Equating coefficients of $x^3$ in $(1)$:
| \(\ds A a^3 + B a^2 - C a^3 + D a^2 - 2 F a^2\) | \(=\) | \(\ds 0\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds A a^3 + \frac {-1} {4 a^5} a^2 - C a^3 + \frac 1 {4 a^5} a^2\) | \(=\) | \(\ds 0\) | substituting for $B$, $D$ and $F$ | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds A - C\) | \(=\) | \(\ds 0\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds A\) | \(=\) | \(\ds C\) | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {-1} {a^6}\) |
Summarising:
| \(\ds A\) | \(=\) | \(\ds \frac {-1} {a^6}\) | ||||||||||||
| \(\ds B\) | \(=\) | \(\ds \frac {-1} {4 a^5}\) | ||||||||||||
| \(\ds C\) | \(=\) | \(\ds \frac {-1} {a^6}\) | ||||||||||||
| \(\ds D\) | \(=\) | \(\ds \frac 1 {4 a^5}\) | ||||||||||||
| \(\ds E\) | \(=\) | \(\ds \frac 2 {a^6}\) | ||||||||||||
| \(\ds F\) | \(=\) | \(\ds 0\) | ||||||||||||
| \(\ds G\) | \(=\) | \(\ds \frac 1 {a^4}\) |
Thus:
| \(\ds \frac 1 {x^3 \paren {x^2 - a^2}^2}\) | \(\equiv\) | \(\ds \dfrac 1 {a^4 x^3} + \dfrac 2 {a^6 x} - \dfrac 1 {a^6 \paren {x + a} } - \dfrac 1 {a^6 \paren {x - a} } - \dfrac 1 {4 a^5 \paren {x + a}^2} + \dfrac 1 {4 a^5 \paren {x - a}^2}\) | ||||||||||||
| \(\ds \) | \(\equiv\) | \(\ds -\frac {\paren {x - a} + \paren {x + a} } {a^6 \paren {x + a} \paren {x - a} } + \frac {\paren {x + a}^2 - \paren {x - a}^2} {4 a^5 \paren {x + a}^2 \paren {x - a}^2} + \dfrac 1 {a^4 x^3} + \dfrac 2 {a^6 x}\) | common denominators | |||||||||||
| \(\ds \) | \(\equiv\) | \(\ds \frac {- 2 x} {a^6 \paren {x^2 - a^2} } + \frac {\paren {x^2 + 2 a x + a^2} - \paren {x^2 - 2 a x + a^2} } {4 a^5 \paren {x^2 - a^2}^2} + \dfrac 1 {a^4 x^3} + \dfrac 2 {a^6 x}\) | simplifying | |||||||||||
| \(\ds \) | \(\equiv\) | \(\ds \frac {-2 x} {a^6 \paren {x^2 - a^2} } + \frac {4 a x} {4 a^5 \paren {x^2 - a^2}^2} + \frac 1 {a^4 x^3} + \frac 2 {a^6 x}\) | simplifying | |||||||||||
| \(\ds \) | \(\equiv\) | \(\ds \frac 1 {a^4 x^3} + \frac 2 {a^6 x} - \frac {2 x} {a^6 \paren {x^2 - a^2} } + \frac x {a^4 \paren {x^2 - a^2}^2}\) | simplifying and rearranging |
Hence the result.
$\blacksquare$