Answer
$$\int \tanh x \, sech^2\, x dx
=\frac{1}{2}\tanh^2 x+c.$$
Work Step by Step
Let $ u=\tanh x $, then $ du=sech^2 x dx $ and hence
$$\int \tanh x \, sech^2\, x dx = \int u du =\frac{1}{2} u^2+c\\
=\frac{1}{2}\tanh^2 x+c.$$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
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: 41
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