Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 5: Integrals - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises 5.5 - Page 295: 37

Answer

$\frac{2}{3}(1+x)^{3/2}-2(1+x)^{1/2}+C$

Work Step by Step

Let u=$1+x $ $x=u-1=>dx=du$ =$\int \frac{x}{\sqrt{1+x}}dx= \int \frac{u-1}{\sqrt{u}}du$ =$\int (u^{1/2}-u^{-1/2})du$ =$\frac{2}{3}u^{3/2}-2u^{1/2}+C$ =$\frac{2}{3}(1+x)^{3/2}-2(1+x)^{1/2}+C$
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