University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 8 - Section 8.5 - Integral Tables and Computer Algebra Systems - Exercises - Page 451: 13



Work Step by Step

$$A=\int\frac{\sqrt{4-x^2}}{x}dx$$ Use Formula 47, which states that $$\int \frac{\sqrt{a^2-x^2}}{x}dx=\sqrt{a^2-x^2}-a\ln\Big|\frac{a+\sqrt{a^2-x^2}}{x}\Big|+C$$ for $a=2$ here: $$A=\sqrt{4-x^2}-2\ln\Big|\frac{2+\sqrt{4-x^2}}{x}\Big|+C$$
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