University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 5 - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises - Page 329: 18



Work Step by Step

$$A=\int\frac{1}{\sqrt{5s+4}}ds$$ We set $u=\sqrt{5s+4}$. Then $$du=\frac{(5s+4)'}{2\sqrt{5s+4}}ds=\frac{5}{2\sqrt{5s+4}}ds$$ That means, $$\frac{1}{\sqrt{5s+4}}ds=\frac{2}{5}du$$ Therefore, $$A=\frac{2}{5}\int du=\frac{2u}{5}+C$$ $$A=\frac{2\sqrt{5s+4}}{5}+C$$
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