Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 6 - Applications of Integration - 6.8 Logarithmic and Exponential - 6.8 Exercises: 23

Answer

$$2{e^{\sqrt x }} + C$$

Work Step by Step

$$\eqalign{ & \int {\frac{{{e^{\sqrt x }}}}{{\sqrt x }}dx} \cr & = \int {{e^{\sqrt x }}\left( {\frac{1}{{\sqrt x }}} \right)dx} \cr & {\text{substitute }}u = \sqrt x ,{\text{ }}du = \frac{1}{{2\sqrt x }}dx \cr & \int {\frac{{{e^{\sqrt x }}}}{{\sqrt x }}dx} = 2\int {{e^u}du} \cr & {\text{find the antiderivative}} \cr & = 2{e^u} + C \cr & {\text{replace back }}u = \sqrt x \cr & = 2{e^{\sqrt x }} + C \cr} $$
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