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: 37


$$\int x \sinh (x^2+1) dx= \frac{1}{2} \cosh (x^2+1)+c.$$

Work Step by Step

Let $ u=x^2+1$, then $ du=2xdx $ and hence $$\int x \sinh (x^2+1) dx= \frac{1}{2}\int \sinh udu = \frac{1}{2}\cosh u+c \\=\frac{1}{2} \cosh (x^2+1)+c.$$
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