Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.9 Hyperbolic Functions - Exercises - Page 384: 54


$$\int\frac{1}{\sqrt{x^2-4}}dx =\frac{1}{2}\cosh^{-1}(x/2)+c.$$

Work Step by Step

Since $\frac{d}{dx} \cosh^{-1}x=\frac{1}{\sqrt{x^2-1}}$, then we have $$\int\frac{1}{\sqrt{x^2-4}}dx=\int\frac{1}{2\sqrt{(x/2)^2-1}}dx=\frac{1}{2}\cosh^{-1}(x/2)+c.$$
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