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: 41



Work Step by Step

Let u=$1-\frac{3}{x^3}=>du=\frac{9}{x^4}dx=>\frac{1}{9}du=\frac{1}{x^4}dx$ =$\int \sqrt{\frac{x^3-3}{x^{11}}}dx$ =$\int \frac{1}{x^4} \sqrt{\frac{x^3-3}{x^{3}}}dx$ =$\int \frac{1}{x^4} \sqrt{1-\frac{3}{x^3}}dx$ =$\int \sqrt{u}\frac{1}{9}du$ =$\frac{1}{9} \int u^{1/2}du$ =$\frac{2}{27}u^{3/2}+C$ =$\frac{2}{27}(1-\frac{3}{x^3})^{3/2}+C$
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