University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 7 - Practice Exercises - Page 421: 28



Work Step by Step

Separate the differentials and integrate. $\int 3(x+1)^2dx=\int (\dfrac{y-1}{y})dy$ or, $3\int (x+1)^2dx=\int (1-\dfrac{1}{y})dy$ This implies that $y-\ln y=\dfrac{3(x+1)^3}{3}+c$ Thus, $y-\ln y=(x+1)^3+c$
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