Calculus: Early Transcendentals 9th Edition

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

Chapter 7 - Review - Exercises - Page 554: 49

Answer

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

Work Step by Step

$$\eqalign{ & \int {\frac{1}{{\sqrt {{e^x} - 4} }}} dx \cr & = \int {\frac{1}{{\sqrt {{{\left( {\sqrt {{e^x}} } \right)}^2} - 4} }}} dx \cr & {\text{Let }}u = \sqrt {{e^x}} ,{\text{ }}du = \frac{{{e^x}}}{{2\sqrt {{e^x}} }}dx,{\text{ }}du = \frac{{\sqrt {{e^x}} }}{2}dx,{\text{ }} \cr & dx = \frac{2}{u}du{\text{ }} \cr & {\text{Substituting we obtain}} \cr & \int {\frac{1}{{\sqrt {{{\left( {\sqrt {{e^x}} } \right)}^2} - 4} }}} dx = \int {\frac{2}{{u\sqrt {{u^2} - 4} }}} du \cr & {\text{ = }}\int {\frac{2}{{u\sqrt {{u^2} - 4} }}} du \cr & {\text{Integrating}} \cr & = {\text{2}}\left( {\frac{1}{2}} \right){\sec ^{ - 1}}\frac{u}{2} + C \cr & {\text{Write in terms of }}x,{\text{ substitute }}\sqrt {{e^x}} {\text{ for }}u \cr & = {\sec ^{ - 1}}\frac{{\sqrt {{e^x}} }}{2} + C \cr} $$
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