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).
Mistake
- $\ds \int \frac {\d x} {\paren {a x + b}^3} = -\frac 1 {2 \paren {a x + b}^2} + C$
As demonstrated in Primitive of $\dfrac 1 {\paren {a x + b}^3}$ the correct expression is in fact:
- $\ds \int \frac {\d x} {\paren {a x + b}^3} = -\frac 1 {2 a \paren {a x + b}^2} + C$