Primitive of Reciprocal of a x + b cubed/Mistake

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Source Work

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

Chapter $14$: Indefinite Integrals:
Integrals involving $a x + b$: $14.73$

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


$\displaystyle \int \frac {\d x} {\paren {a x + b}^3} = -\frac 1 {2 \paren {a x + b}^2} + C$

As demonstrated in Primitive of $\dfrac 1 {\paren {a x + b}^3}$ the correct expression is in fact:

$\displaystyle \int \frac {\d x} {\paren {a x + b}^3} = -\frac 1 {2 a \paren {a x + b}^2} + C$