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: 21


$$\int (4t^{-3}-12t^{-4})dt =-2t^{-2}+4t^{-3}+c.$$

Work Step by Step

We have $$\int (4t^{-3}-12t^{-4})dt =-\frac{4}{2}t^{-2}+\frac{12}{3}t^{-3}+c=-2t^{-2}+4t^{-3}+c.$$
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