Answer
$$\frac{1}{2\ln 3}3^{x^2}+c$$
Work Step by Step
Let $u=x^2$, then $du=2xdx$ and hence we get
$$\int x3^{x^2}dx=\frac{1}{2}\int 3^u du \\
=\frac{1}{2\ln 3}3^u+c=\frac{1}{2\ln 3}3^{x^2}+c$$
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 - 7.8 Inverse Trigonometric Functions - Exercises - Page 375: 105
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