Answer
$\frac{-2}{1+\sqrt{x}}+C$
Work Step by Step
Let u=1+$\sqrt{x} =>du=\frac{1}{2\sqrt{x}}dx=>2du=\frac{1}{\sqrt{x}}dx$
=$\int \frac{1}{\sqrt{x}(1+\sqrt{x})^2}dx$
=$\int \frac{2du}{u^2}=\frac{-2}{u}+C$
=$\frac{-2}{1+\sqrt{x}}+C$
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
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
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