Primitive of Reciprocal of x cubed by a x + b squared/Partial Fraction Expansion
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Lemma for Primitive of $\dfrac 1 {x^3 \paren {a x + b}^2}$
- $\dfrac 1 {x^3 \paren {a x + b}^2} \equiv \dfrac {3 a^2} {b^4 x} - \dfrac {2 a} {b^3 x^2} + \dfrac 1 {b^2 x^3} - \dfrac {3 a^3} {b^4 \paren {a x + b} } - \dfrac {a^3} {b^3 \paren {a x + b}^2}$
Proof
| \(\ds \dfrac 1 {x^3 \paren {a x + b}^2}\) | \(\equiv\) | \(\ds \dfrac A x + \dfrac B {x^2} + \dfrac C {x^3} + \dfrac D {a x + b} + \dfrac E {\paren {a x + b}^2}\) | ||||||||||||
| \(\text {(1)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds 1\) | \(\equiv\) | \(\ds A x^2 \paren {a x + b}^2 + B x \paren {a x + b}^2 + C \paren {a x + b}^2 + D x^3 \paren {a x + b} + E x^3\) | multiplying through by $x^3 \paren {a x + b}^2$ | |||||||||
| \(\ds \) | \(\equiv\) | \(\ds A a^2 x^4 + 2 A a b x^3 + A b^2 x^2\) | multiplying everything out | |||||||||||
| \(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds B a^2 x^3 + 2 B a b x^2 + B b^2 x\) | (tedious though this is, it helps to | ||||||||||
| \(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds C a^2 x^2 + 2 C a b x + C b^2\) | identify the equal indices) | ||||||||||
| \(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds D a x^4 + D b x^3 + E x^3\) |
Setting $a x + b = 0$ in $(1)$:
| \(\ds a x + b\) | \(=\) | \(\ds 0\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds -\frac b a\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds E \paren {-\frac b a}^3\) | \(=\) | \(\ds 1\) | substituting for $x$ in $(1)$: terms in $a x + b$ are all $0$ | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds E\) | \(=\) | \(\ds -\frac {a^3} {b^3}\) |
Equating constants in $(1)$:
| \(\ds 1\) | \(=\) | \(\ds C b^2\) | ||||||||||||
| \(\text {(2)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds C\) | \(=\) | \(\ds \frac 1 {b^2}\) |
Equating $1$st powers of $x$ in $(1)$:
| \(\ds 0\) | \(=\) | \(\ds 2 C a b + B b^2\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \frac {2 a b} {b^2}\) | \(=\) | \(\ds -B b^2\) | subtituting for $C$ from $(2)$ | ||||||||||
| \(\text {(3)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds B\) | \(=\) | \(\ds -\frac {2 a} {b^3}\) |
Equating $2$nd powers of $x$ in $(1)$:
| \(\ds 0\) | \(=\) | \(\ds A b^2 + 2 B a b + C a^2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds A b^2 + 2 \paren {-\frac {2 a} {b^3} } a b + \paren {\frac 1 {b^2} } a^2\) | substituting for $B$ and $C$ from $(2)$ and $(3)$ | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds A b^2\) | \(=\) | \(\ds \frac {4 a^2} {b^2} - \frac {a^2} {b^2}\) | rearranging | ||||||||||
| \(\text {(4)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds A\) | \(=\) | \(\ds \frac {3 a^2} {b^4}\) | simplifying |
Equating $4$th powers of $x$:
| \(\ds 0\) | \(=\) | \(\ds A a^2 + D a\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds D\) | \(=\) | \(\ds \frac {3 a^3} {b^4}\) | substituting for $A$ from $(3)$ and simplifying |
Summarising:
| \(\ds A\) | \(=\) | \(\ds \frac {3 a^2} {b^4}\) | ||||||||||||
| \(\ds B\) | \(=\) | \(\ds -\frac {2 a} {b^3}\) | ||||||||||||
| \(\ds C\) | \(=\) | \(\ds \frac 1 {b^2}\) | ||||||||||||
| \(\ds D\) | \(=\) | \(\ds \frac {3 a^3} {b^4}\) | ||||||||||||
| \(\ds E\) | \(=\) | \(\ds -\frac {a^3} {b^3}\) |
Hence the result.
$\blacksquare$