Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 7 - Review - Exercises - Page 553: 24

Answer

$\frac{{{2^{\sqrt x + 1}}}}{{\ln 2}} + C$

Work Step by Step

$$\eqalign{ & \int {\frac{{{2^{\sqrt x }}}}{{\sqrt x }}} dx \cr & = \int {{2^{\sqrt x }}\left( {\frac{1}{{\sqrt x }}} \right)} dx \cr & {\text{Let }}u = \sqrt x ,{\text{ }}du = \frac{1}{{2\sqrt x }}dx,{\text{ }}2du = \frac{1}{{\sqrt x }}dx \cr & {\text{Applying the substitution}}{\text{, we obtain}} \cr & \int {{2^{\sqrt x }}\left( {\frac{1}{{\sqrt x }}} \right)} dx = \int {{2^u}\left( 2 \right)} du \cr & = 2\int {{2^u}} du \cr & {\text{Integrating}} \cr & = 2\left( {\frac{{{2^u}}}{{\ln 2}}} \right) + C \cr & = \frac{{{2^{u + 1}}}}{{\ln 2}} + C \cr & {\text{Write in terms of }}x,{\text{ substitute }}\sqrt x {\text{ for }}u \cr & = \frac{{{2^{\sqrt x + 1}}}}{{\ln 2}} + C \cr} $$
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