Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.2 Exercises - Page 334: 27



Work Step by Step

$\int\frac{1}{1+\sqrt{2x}}dx$ let $\sqrt{2x} = (2x)^{\frac{1}{2}} = u$ $du=\frac{1}{2}(2x)^{-\frac{1}{2}}(2)dx$ $du=\frac{1}{\sqrt{2x}}dx$ $du=\frac{1}{u}dx$ $dx=u$ $du$ $\int\frac{1}{1+\sqrt{2x}}dx$ $=\int\frac{u}{1+u}du$ $=\int\frac{1+u-1}{1+u}du$ $=\int(1-\frac{1}{1+u})du$ $=u-ln|1+u|+C$ $=\sqrt{2x}-ln|1+\sqrt{2x}|+C$
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