Calculus (3rd Edition)

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

Chapter 7 - Exponential Functions - Chapter Review Exercises - Page 387: 66


$\tan^{-1}(\tanh t)+c $

Work Step by Step

Since $ u=\tanh t $, then $ du=\operatorname{sech}^2 tdt $. Now, we have $$\int \frac{dt}{\cosh^2 +\sinh^2t }=\int \frac{\operatorname{sech}^2tdt}{1 +\tanh^2t }=\int \frac{1}{1+u^2}du\\ =\tan^{-1}u+c=\tan^{-1}(\tanh t)+c.$$
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