Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.4 Exercises: 93

Answer

$\frac{1}{2}e^{2x-1}+C$

Work Step by Step

$\int e^{2x-1}dx$ let $u=2x-1$ $2dx=du$ $dx=\frac{1}{2}du$ $\int e^{2x-1}dx$ $=\frac{1}{2} \int e^u du$ $=\frac{1}{2}e^u+C$ $=\frac{1}{2}e^{2x-1}+C$
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