Answer
$\int(5t^3-t^{-3})dt = \frac{5}{4}t^4 +\frac{1}{2t^2}+C$
Work Step by Step
$\int(5t^3-t^{-3})dt = 5\times\frac{1}{4}t^4 -\frac{1}{-2}t^{-2}+C = \frac{5}{4}t^4 +\frac{1}{2t^2}+C$
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 - 5.3 The Indefinite Integral - Exercises - Page 252: 14
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