Primitive of Reciprocal of a x + b cubed/Mistake

Source Work

 * 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).

Mistake

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

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


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