Calculus: Early Transcendentals (2nd Edition)

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

Chapter 4 - Applications of the Derivative - 4.9 Antiderivatives - 4.9 Exercises: 20

Answer

$H(y) = ln(y)+c$

Work Step by Step

$h(y)=y^{-1}$ $\int h(y)dy = \int y^{-1}$ $H(y) = ln(y)+c$
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