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