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: 16


$\frac{1}{40}(4x-1)^{5/2}+\frac{1}{24}(4x-1)^{3/2} +c $

Work Step by Step

Since $ u=4x-1$, then $ du=4dx $ and hence, $$ \int x \sqrt{4x-1} dx=\frac{1}{16}\int (u+1)u^{1/2} d u=\frac{1}{16}\int u^{3/2}+u^{1/2} d u\\ =\frac{1}{16}\frac{2}{5}u^{5/2}+\frac{1}{16}\frac{2}{3}u^{3/2}+c=\frac{1}{40}(4x-1)^{5/2}+\frac{1}{24}(4x-1)^{3/2} +c $$
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