Calculus (3rd Edition)

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

Chapter 5 - The Integral - 5.3 The Indefinite Integral - Exercises - Page 252: 14


$\int(5t^3-t^{-3})dt = \frac{5}{4}t^4 +\frac{1}{2t^2}+C$

Work Step by Step

$\int(5t^3-t^{-3})dt = 5\times\frac{1}{4}t^4 -\frac{1}{-2}t^{-2}+C = \frac{5}{4}t^4 +\frac{1}{2t^2}+C$
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