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: 42



Work Step by Step

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