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: 28



Work Step by Step

$\int\frac{1}{1+\sqrt{3x}}dx$ let $\sqrt{3x}=(3x^{\frac{1}{2}})=u$ $du=\frac{1}{2}(3x)^{-\frac{1}{2}}(3)dx$ $du=\frac{3}{2}(3x)^{-\frac{1}{2}} dx=\frac{3}{2\sqrt{3x}}dx$ $du= \frac{3}{2u}dx$ $(2u)du=3dx$ $dx=\frac{(2u)du}{3}$ $\int\frac{1}{1+\sqrt{3x}}dx$ $=\int\frac{2u}{3(1+u}du$ $=\frac{2}{3}\int\frac{u}{1+u}du$ $=\frac{2}{3}\int\frac{1+u-1}{1+u}du$ $=\frac{2}{3}\int(1-\frac{1}{1+u})du$ $=\frac{2}{3}(u-ln|1+u|)+C$ $=\frac{2}{3}(\sqrt{3x}-ln|(1+\sqrt{3x})|)+C$
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