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: 31



Work Step by Step

Since, we have $\int (\dfrac{1}{x^2}-x^2-\dfrac{1}{3}) dx=-\dfrac{1}{x}-\dfrac{x^3}{3}-\dfrac{x}{3}+C$ and $\dfrac{d}{dx}(-\dfrac{1}{x}-\dfrac{x^3}{3}-\dfrac{x}{3}+C)=\dfrac{1}{x^2}-x^2-\dfrac{1}{3}$
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