Primitive of Reciprocal of p x + q by Root of a x + b/Lemma

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:

$u = \sqrt {a x + b}$

Let:

$\ds u$

$=$

$\ds \sqrt {a x + b}$

$\ds \leadsto \ \ $

$\ds x$

$=$

$\ds \frac {u^2 - b} a$

and:

$\ds \dfrac {\d u} {\d x}$

$=$

$\ds \dfrac a {2 \sqrt {a x + b} }$

$\ds \leadsto \ \ $

$\ds \dfrac {\d x} {\d u}$

$=$

$\ds \dfrac {2 \sqrt {a x + b} } a$

Then:

$\ds \int \frac {\d x} {\paren {p x + q} \sqrt {a x + b} }$

$=$

$\ds \int \frac 1 {\paren {p \paren {\frac {u^2 - b} a} + q} \sqrt {a x + b} } \dfrac {2 \sqrt {a x + b} } a \rd u$

$\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

$\blacksquare$