University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 7 - Section 7.3 - Hyperbolic Functions - Exercises - Page 418: 64



Work Step by Step

We have $$\coth^{-1}x=\frac{1}{2}\ln\frac{x+1}{x-1}$$ for $|x|\gt1$ Therefore, $$\coth^{-1}\Big(\frac{5}{4}\Big)=\frac{1}{2}\ln\frac{5/4+1}{5/4-1}$$ $$=\frac{1}{2}\ln\frac{9/4}{1/4}=\frac{1}{2}\ln9$$ $$=\ln9^{1/2}=\ln3$$
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