University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 4 - Section 4.8 - Antiderivatives - Exercises - Page 272: 76



Work Step by Step

Simplify $\dfrac{d(\dfrac{x}{x+1}+C)}{dx}$ Thus, we have $\dfrac{d(\dfrac{x}{x+1}+C)}{dx}=\dfrac{d(1-\dfrac{1}{x+1}+C)}{dx}=(x+1)^{-2}=\dfrac{1}{(x+1)^2}$
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