Answer
$$\int \frac{8}{x^3} dx =-\frac{4}{x^{2}}+c.$$
Work Step by Step
We have
$$\int \frac{8}{x^3} dx =\int 8x^{-3} dx =-\frac{8}{2}x^{-2}+c=-\frac{4}{x^{2}}+c.$$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
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
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