Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 5 - Integration - Review Exercises - Page 395: 22


$\frac{1}{3}\ln (y^3+27)+c$

Work Step by Step

$\int\frac{y^2}{y^3+27}dy$ Use u-substitution... $u=y^3+27$ $dy=\frac{du}{3y^2}$ $\int\frac{y^2}{u}\frac{du}{3y^2}=\frac{1}{3}\int\frac{1}{u}du$ $=\frac{1}{3}\ln u+c$ substitute out u... $\frac{1}{3}\ln (y^3+27)+c$
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