Primitive of Reciprocal of a x + b squared by p x + q/Partial Fraction Expansion
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Lemma for Primitive of Reciprocal of a x + b squared by p x + q
- $\dfrac 1 {\paren {a x + b}^2 \paren {p x + q} } \equiv \dfrac 1 {b p - a q} \paren {\dfrac {-a p} {\paren {b p - a q} \paren {a x + b} } + \dfrac {-a} {\paren {a x + b}^2} + \dfrac {p^2} {\paren {b p - a q} \paren {p x + q} } }$
Proof
| \(\ds \frac 1 {\paren {a x + b}^2 \paren {p x + q} }\) | \(\equiv\) | \(\ds \frac A {a x + b} + \frac B {\paren {a x + b}^2} + \frac C {p x + q}\) | ||||||||||||
| \(\text {(1)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds 1\) | \(\equiv\) | \(\ds A \paren {a x + b} \paren {p x + q} + B \paren {p x + q} + C \paren {a x + b}^2\) | multiplying through by $\paren {a x + b}^2 \paren {p x + q}$ |
Setting $a x + b = 0$ in $(1)$:
| \(\ds a x + b\) | \(=\) | \(\ds 0\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds -\frac b a\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds B \paren {p \paren {-\frac b a} + q}\) | \(=\) | \(\ds 1\) | substituting for $x$ in $(1)$: term in $a x + b$ is $0$ | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds B\) | \(=\) | \(\ds \frac {-a} {b p - a q}\) |
Setting $p x + q = 0$ in $(1)$:
| \(\ds p x + q\) | \(=\) | \(\ds 0\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds -\frac q p\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds C \paren {a \paren {-\frac q p} + b}^2\) | \(=\) | \(\ds 1\) | substituting for $x$ in $(1)$: term in $a x + b$ is $0$ | ||||||||||
| \(\text {(2)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds C\) | \(=\) | \(\ds \frac {p^2} {\paren {b p - a q}^2}\) |
Equating $2$nd powers of $x$ in $(1)$:
| \(\ds 0\) | \(=\) | \(\ds A a p + C a^2\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds A\) | \(=\) | \(\ds -C \frac a p\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds A\) | \(=\) | \(\ds -\frac {p^2} {\paren {b p - a q}^2} \frac a p\) | substituting for $C$ from $(2)$ | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds A\) | \(=\) | \(\ds \frac {-a p} {\paren {b p - a q}^2}\) | simplifying |
Summarising:
| \(\ds A\) | \(=\) | \(\ds \frac {-a p} {\paren {b p - a q}^2}\) | ||||||||||||
| \(\ds B\) | \(=\) | \(\ds \frac {-a} {b p - a q}\) | ||||||||||||
| \(\ds C\) | \(=\) | \(\ds \frac {p^2} {\paren {b p - a q}^2}\) |
Hence the result.
$\blacksquare$