Thomas' Calculus 13th Edition

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

Chapter 8: Techniques of Integration - Section 8.1 - Using Basic Integration Formulas - Exercises 8.1 - Page 448: 6

Answer

$2 \ln|\sqrt {x}-1|+C$

Work Step by Step

$\int \frac{dx}{x-\sqrt x}=\int\frac{dx}{\sqrt x(\sqrt x-1)}$ Put $t= \sqrt {x}-1$ Then $dx= 2\sqrt x dt$ Therefore, $\int \frac{dx}{x-\sqrt x}= 2\int \frac{1}{t}dt$ $= 2\times\ln|t|+C= 2 \ln|\sqrt {x}-1|+C$
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