Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 6 - Exponential, Logarithmic, And Inverse Trigonometric Functions - 6.3 Derivatives Of Inverse Functions; Derivatives And Integrals Involving Exponential Functions - Exercises Set 6.3 - Page 433: 65



Work Step by Step

Using a u-substitution of $u=2x$ leaves a $du=2dx$, meaning that $\frac{1}{2}du=dx$. This yields $$\frac{1}{2}\int{e^{u}}du=\frac{1}{2}e^{u}+C$$ Changing back to terms of $u$ yields $\frac{1}{2}e^{2x}+C.$
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