Primitive of Reciprocal of x cubed by a x + b cubed/Mistake

From ProofWiki
Jump to navigation Jump to search

Source Work

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

Chapter $14$: Indefinite Integrals
Integrals Involving $a x + b$: $14.79$


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


Mistake

$\displaystyle \int \frac {\d x} {x^3 \paren {a x + b}^3} = \frac {a^4 x^2} {2 b^5 \paren {a x + b}^2} - \frac {4 a^3 x} {b^5 \paren {a x + b} } - \frac {\paren {a x + b}^2} {2 b^5 x^2} + \frac {6 a^2} {b^5} \ln \size {\frac x {a x + b} } + C$


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

$\displaystyle \int \frac {\d x} {x^3 \paren {a x + b}^3} = \frac {a^4 x^2} {2 b^5 \paren {a x + b}^2} - \frac {4 a^3 x} {b^5 \paren {a x + b} } - \frac {\paren {a x + b}^2} {2 b^5 x^2} + \frac {4 a} {b^4 x} + \frac {6 a^2} {b^5} \ln \size {\frac x {a x + b} } + C$