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

 * $\displaystyle \int \frac {\mathrm d x} {\sqrt{\left({a x + b}\right) \left({p x + q}\right)} } = \begin{cases}

\dfrac 2 {\sqrt {a p} } \ln \left({\sqrt {p \left({a x + b}\right)} + \sqrt {a \left({p x + q}\right)} }\right) + C & : \dfrac {b p - a q} p > 0 \\ \dfrac 2 {-\sqrt {a p} } \arctan \sqrt {\dfrac {-p \left({a x + b}\right)} {a \left({p x + q}\right)} } + C & :\dfrac {b p - a q} p < 0 \\ \end{cases}$

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


 * $\displaystyle \int \frac {\mathrm d x} {\sqrt{\left({a x + b}\right) \left({p x + q}\right)} } = \begin{cases}

\dfrac 2 {\sqrt {a p} } \ln \left({\sqrt {p \left({a x + b}\right)} + \sqrt {a \left({p x + q}\right)} }\right) + C & : \dfrac {b p - a q} p > 0 \\ \dfrac 2 {\sqrt {a p} } \sinh^{-1} \sqrt {\dfrac {p \left({a x + b}\right)} {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.