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: 16

Answer

$$\int\frac{dx}{\sqrt{5x+8}}=\frac{2\sqrt{5x+8}}{5}+C$$

Work Step by Step

$$A=\int\frac{dx}{\sqrt{5x+8}}$$ a) We set $u=5x+8$. Then $$du=5dx$$ That means, $$dx=\frac{1}{5}du$$ Therefore, $$A=\frac{1}{5}\int\frac{du}{\sqrt u}=\frac{1}{5}\int u^{-1/2}du$$ $$A=\frac{1}{5}\times\frac{u^{1/2}}{\frac{1}{2}}+C=\frac{1}{5}\times2\sqrt u+C=\frac{2\sqrt u}{5}+C$$ $$A=\frac{2\sqrt{5x+8}}{5}+C$$ b) We set $u=\sqrt{5x+8}$. Then $$du=\frac{(5x+8)'}{2\sqrt{5x+8}}dx=\frac{5}{2u}dx$$ That means, $$dx=\frac{2u}{5}du$$ Therefore, $$A=\int\frac{\frac{2u}{5}}{u}du=\int\frac{2u}{5u}du=\int\frac{2}{5}du$$ $$A=\frac{2u}{5}+C$$ $$A=\frac{2\sqrt{5x+8}}{5}+C$$
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