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: 35



Work Step by Step

Given: $\int (\sqrt x+\sqrt [3] x) dx$ or, $\int (\sqrt x+\sqrt [3] x) dx=\int (x^{1/2}+ x^{1/3}) dx$ We have: $\int x^n dx=\dfrac{x^{n+1}}{n+1}+C$ Thus, $\int (x^{1/2}+ x^{1/3}) dx=\dfrac{x^{3/2}}{3/2}+\dfrac{x^{4/3}}{4/3}+C=\dfrac{2}{3}x^{3/2}+\dfrac{3}{4}x^{4/3}+C$
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