Primitive of Reciprocal of x squared by a x + b cubed/Partial Fraction Expansion
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Lemma for Primitive of $\dfrac 1 {x^2 \paren {a x + b}^3}$
- $\dfrac 1 {x^2 \paren {a x + b}^3} \equiv \dfrac {-3 a} {b^4 x} + \dfrac 1 {b^3 x^2} + \dfrac {3 a^2} {b^4 \paren {a x + b} } + \dfrac {2 a^2} {b^3 \paren {a x + b}^2} + \dfrac {a^2} {b^2 \paren {a x + b}^3}$
Proof
| \(\ds \dfrac 1 {x^2 \paren {a x + b}^3}\) | \(\equiv\) | \(\ds \dfrac A x + \dfrac B {x^2} + \dfrac C {a x + b} + \dfrac D {\paren {a x + b}^2} + \dfrac E {\paren {a x + b}^3}\) | ||||||||||||
| \(\text {(1)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds 1\) | \(\equiv\) | \(\ds A x \paren {a x + b}^3 + B \paren {a x + b}^3 + C x^2 \paren {a x + b}^2 + D x^2 \paren {a x + b} + E x^2\) | multiplying through by $x^2 \paren {a x + b}^3$ | |||||||||
| \(\ds \) | \(\equiv\) | \(\ds A a^3 x^4 + 3 A a^2 b x^3 + 3 A a b^2 x^2 + A b^3 x\) | multiplying everything out | |||||||||||
| \(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds B a^3 x^3 + 3 B a^2 b x^2 + 3 B a b^2 x + B b^3\) | |||||||||||
| \(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds C a^2 x^4 + 2 C a b x^3 + C b^2 x^2 + D a x^3 + D b x^2 + E x^2\) |
Setting $a x + b = 0$ in $(1)$:
| \(\ds a x + b\) | \(=\) | \(\ds 0\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds -\frac b a\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds E \paren {-\frac b a}^2\) | \(=\) | \(\ds 1\) | substituting for $x$ in $(1)$: terms in $a x + b$ are all $0$ | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds E\) | \(=\) | \(\ds \frac {a^2} {b^2}\) |
Equating constants in $(1)$:
| \(\ds 1\) | \(=\) | \(\ds B b^3\) | ||||||||||||
| \(\text {(2)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds B\) | \(=\) | \(\ds \frac 1 {b^3}\) |
Equating $1$st powers of $x$ in $(1)$:
| \(\ds 0\) | \(=\) | \(\ds A b^3 + 3 B a b^2\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds A b^3\) | \(=\) | \(\ds -3 \frac 1 {b^3} a b^2\) | substituting for $B$ from $(2)$ | ||||||||||
| \(\text {(3)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds A\) | \(=\) | \(\ds \frac {-3 a} {b^4}\) | simplifying |
Equating $4$th powers of $x$ in $(1)$:
| \(\ds 0\) | \(=\) | \(\ds A a^3 + C a^2\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds C a^2\) | \(=\) | \(\ds \frac {3 a} {b^4} a^3\) | substituting for $A$ from $(3)$ | ||||||||||
| \(\text {(4)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds C\) | \(=\) | \(\ds \frac {3 a^2} {b^4}\) | simplifying |
Equating $3$rd powers of $x$ in $(1)$:
| \(\ds 0\) | \(=\) | \(\ds 3 A a^2 b + B a^3 + 2 C a b + D a\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds D\) | \(=\) | \(\ds -3 A a b - B a^2 - 2 C b\) | rearranging | ||||||||||
| \(\ds \) | \(=\) | \(\ds -3 a b \frac {-3 a} {b^4} - \frac 1 {b^3} a^2 - 2 b \frac {3 a^2} {b^4}\) | substituting for $A$ from $(3)$, $B$ from $(2)$ and $C$ from $(4)$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {9 a^2} {b^3} - \frac {a^2} {b^3} - \frac {6 a^2} {b^3}\) | simplifying | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {2 a^2} {b^3}\) | simplifying |
Summarising:
| \(\ds A\) | \(=\) | \(\ds \frac {-3 a} {b^4}\) | ||||||||||||
| \(\ds B\) | \(=\) | \(\ds \frac 1 {b^3}\) | ||||||||||||
| \(\ds C\) | \(=\) | \(\ds \frac {3 a^2} {b^4}\) | ||||||||||||
| \(\ds D\) | \(=\) | \(\ds \frac {2 a^2} {b^3}\) | ||||||||||||
| \(\ds E\) | \(=\) | \(\ds \frac {a^2} {b^2}\) |
Hence the result.
$\blacksquare$