University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Practice Exercises - Page 343: 61


$$\int x3^{x^2}dx=\frac{3^{x^2}}{2\ln3}+C$$

Work Step by Step

$$A=\int x3^{x^2}dx$$ We set $a=x^2$, which means $$da=2xdx$$ $$xdx=\frac{1}{2}da$$ Therefore, $$A=\frac{1}{2}\int3^ada$$ $$A=\frac{1}{2}\times\frac{3^a}{\ln3}+C$$ $$A=\frac{3^a}{2\ln3}+C$$ $$A=\frac{3^{x^2}}{2\ln3}+C$$
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