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

## Source Work

Chapter $14$: Indefinite Integrals
Integrals involving $\sqrt {a x + b}$ and $\sqrt{p x + q}$: $14.120$

This mistake can be seen in the edition as published by Schaum: ISBN 0-07-060224-7 (unknown printing).

## Mistake

$\ds \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } } = \begin {cases} \dfrac 2 {\sqrt {a p} } \map \ln {\sqrt {p \paren {a x + b} } + \sqrt {a \paren {p x + q} } } + C & : \dfrac {b p - a q} p > 0 \\ \dfrac 2 {-\sqrt {a p} } \arctan \sqrt {\dfrac {-p \paren {a x + b} } {a \paren {p x + q} } } + C & :\dfrac {b p - a q} p < 0 \\ \end{cases}$

As demonstrated in Primitive of Reciprocal of Root of $\sqrt {\paren {a x + b} \paren {p x + q} }$ the correct expression is in fact:

$\ds \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } } = \begin {cases} \dfrac 2 {\sqrt {a p} } \map \ln {\sqrt {p \paren {a x + b} } + \sqrt {a \paren {p x + q} } } + C & : \dfrac {b p - a q} p > 0 \\ \dfrac 2 {\sqrt {a p} } \sinh^{-1} \sqrt {\dfrac {p \paren {a x + b} } {b p - a q} } + C & : \dfrac {b p - a q} p < 0 \\ \end{cases}$

It is suspected that the mistake may have arisen through omitting to take into account a square root.