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: 21

Answer

$\frac{-2}{1+\sqrt{x}}+C$

Work Step by Step

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