Answer
$$\int x \sinh (x^2+1) dx= \frac{1}{2} \cosh (x^2+1)+c.$$
Work Step by Step
Let $ u=x^2+1$, then $ du=2xdx $ and hence
$$\int x \sinh (x^2+1) dx= \frac{1}{2}\int \sinh udu = \frac{1}{2}\cosh u+c
\\=\frac{1}{2} \cosh (x^2+1)+c.$$
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: 37
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