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: 25


$$\int (y+2)^4dy = \frac{1}{5}(y+2)^5+c.$$

Work Step by Step

Let $ u=y+2$, then $ du=dy $ and hence we have $$\int (y+2)^4dy =\in u^4fu =\frac{1}{5}u^5+c= \frac{1}{5}(y+2)^5+c $$
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