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: 38



Work Step by Step

Let u=$1-\frac{1}{x}=>du=\frac{1}{x^2}dx$ =$\frac{\sqrt{x-1}}{\sqrt{x^5}}dx$ =$\int \sqrt{\frac{x-1}{x}}dx$ =$\int \frac{1}{x^2}\sqrt{1-\frac{1}{x}}dx$ =$\int \sqrt{u}du$ =$\int u^{1/2}du$ =$\frac{2}{3}u^{3/2}+C$ =$\frac{2}{3}(1-\frac{1}{x})^{3/2}+C$
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