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

Source Work

1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables

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} } } \\ \dfrac 2 {\sqrt {-a p} } \tan^{-1} \sqrt {\dfrac {-p \paren {a x + b} } {a \paren {p x + q} } } \end{cases}$

Correction

It is important to give the conditions under which the primitives are valid.

It is also important to specify that the argument of the logarithm function is absolute:

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

\dfrac 2 {\sqrt {a p} } \ln \size {\sqrt {p \paren {a x + b} } + \sqrt {a \paren {p x + q} } } & : a p > 0 \\ \dfrac 2 {\sqrt {-a p} } \tan^{-1} \sqrt {\dfrac {-p \paren {a x + b} } {a \paren {p x + q} } } & : a p < 0 \end{cases}$

Sources

• 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\sqrt {a x + b}$ and $\sqrt{p x + q}$: $14.120$