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: 42


$$\int \frac{\cosh x}{3\sinh x +4} dx =\frac{1}{3} \ln (3\sinh x +4)+c $$

Work Step by Step

Let $ u=3\sinh x +4$, then $ du=3\cosh x dx $ and hence $$\int \frac{\cosh x}{3\sinh x +4} dx = \frac{1}{3}\int \frac{du}{u}=\frac{1}{3} \ln u+c\\ =\frac{1}{3} \ln (3\sinh x +4)+c $$
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