Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 5 - The Integral - 5.7 Substitution Method - Exercises - Page 275: 14



Work Step by Step

Substituting u=4x-1 so that $dx=\frac{du}{4}$ We get $\int \sqrt {4x-1}dx=\int u^{\frac{1}{2}}\frac{du}{4}$ $=\frac{1}{4}\int u^{\frac{1}{2}}du=\frac{1}{4}\times\frac{2}{3}u^{\frac{3}{2}}+C$ $=\frac{1}{6}u^{\frac{3}{2}}+C$ $=\frac{1}{6}(4x-1)^{\frac{3}{2}}+C$
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