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: 40


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

Work Step by Step

We use the u-substitution: $u=1-\frac{1}{x^2}$ Derive: $du=\frac{2}{x^3}dx$ $\frac{1}{2}du=\frac{1}{x^3}dx $ Substitute: $\int \frac{1}{x^3}\sqrt{\frac{x^2 -1}{x^2}}dx $ =$\int \frac{1}{x^3}\sqrt{1-\frac{1}{x^2}}dx $ =$\int \sqrt{u}\frac{1}{2}du $ =$\frac{1}{2}\int u^{1/2}du $ =$\frac{1}{3}u^{3/2}+C $ =$\frac{1}{3}(1-\frac{1}{x^2})^{3/2}+C $
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