Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 9 - Problems Plus: 2

Answer

$$C\sqrt {2x-1}e^{x}$$

Work Step by Step

$$(e^{x^2} g(x))' = e^{x^2} 2x g(x) + e^{x^2} g'(x) = e^{x^2} 2x g'(x)$$ $$2x g(x) = (2x-1) g'(x)$$ $$\int{g'(x)/g(x)dx} = \int{2x/(2x-1)dx}$$ $$\ln{g(x)} = x+\frac{1}{2} \ln{(2x-1)} +C$$ $$g(x) = C\sqrt {2x-1}e^{x}$$
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