Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 5 - The Integral - Chapter Review Exercises - Page 278: 29


$$\int \frac{8}{x^3} dx =-\frac{4}{x^{2}}+c.$$

Work Step by Step

We have $$\int \frac{8}{x^3} dx =\int 8x^{-3} dx =-\frac{8}{2}x^{-2}+c=-\frac{4}{x^{2}}+c.$$
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