Primitive of Reciprocal of p x + q by Root of a x + b/Lemma
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Lemma for Primitive of $\frac 1 {\paren {p x + q} \sqrt {a x + b} }$
Let $a, b, p, q \in \R$ such that $a p \ne b q$ and such that $p \ne 0$.
Then:
- $\ds \int \frac {\d x} {\paren {p x + q} \sqrt {a x + b} } = \dfrac 2 p \int \frac {\d u} {u^2 - \paren {\dfrac {b p - a q} p} }$
where:
- $u = \sqrt {a x + b}$
Proof
Let:
\(\ds u\) | \(=\) | \(\ds \sqrt {a x + b}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds \frac {u^2 - b} a\) |
and:
\(\ds \dfrac {\d u} {\d x}\) | \(=\) | \(\ds \dfrac a {2 \sqrt {a x + b} }\) | Power Rule for Derivatives, Chain Rule for Derivatives | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac {\d x} {\d u}\) | \(=\) | \(\ds \dfrac {2 \sqrt {a x + b} } a\) | Derivative of Inverse Function |
Then:
\(\ds \int \frac {\d x} {\paren {p x + q} \sqrt {a x + b} }\) | \(=\) | \(\ds \int \frac 1 {\paren {p \paren {\frac {u^2 - b} a} + q} \sqrt {a x + b} } \dfrac {2 \sqrt {a x + b} } a \rd u\) | Integration by Substitution | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 2 a \int \frac {\d u} {\paren {\dfrac {p \paren {u^2 - b} + a q} a} }\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 2 p \int \frac {\d u} {u^2 - \dfrac {b p - a q} p}\) | simplifying |
$\blacksquare$