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