University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises - Page 330: 40



Work Step by Step

$$A=\int\frac{1}{x^3}\sqrt{\frac{x^2-1}{x^2}}dx$$ $$A=\int\frac{1}{x^3}\sqrt{1-\frac{1}{x^2}}dx$$ We set $u=1-\frac{1}{x^2}$ Then $$du=\Big(0-\frac{(1)'x^2-1(x^2)'}{x^4}\Big)dx=-\Big(\frac{0-2x}{x^4}\Big)dx$$ $$du=-\Big(-\frac{2}{x^3}\Big)dx=\frac{2}{x^3}dx$$ That means $$\frac{1}{x^3}dx=\frac{1}{2}du$$ Therefore, $$A=\frac{1}{2}\int\sqrt udu=\frac{1}{2}\int u^{1/2}du$$ $$A=\frac{1}{2}\times\frac{u^{3/2}}{\frac{3}{2}}+C=\frac{u^{3/2}}{3}+C$$ $$A=\frac{1}{3}\Big(1-\frac{1}{x^2}\Big)^{3/2}+C$$
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