Calculus: Early Transcendentals (2nd Edition)

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

Chapter 5 - Integration - 5.5 Substitution Rule - 5.5 Exercises: 36

Answer

$$\ln \left( {{e^x} + {e^{ - x}}} \right) + C$$

Work Step by Step

$$\eqalign{ & \int {\frac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}}dx} \cr & {\text{substitute }}u = {e^x} + {e^{ - x}}, \cr & {\text{ }}du = \left( {{e^x} - {e^{ - x}}} \right)dx \cr & \int {\frac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}}dx} = \int {\frac{{du}}{u}} \cr & {\text{find the antiderivative use the logarithmic rule}} \cr & = \ln \left| u \right| + C \cr & {\text{ with}}\,\,\,u = {e^x} + {e^{ - x}} \cr & = \ln \left| {{e^x} + {e^{ - x}}} \right| + C \cr & = \ln \left( {{e^x} + {e^{ - x}}} \right) + C \cr} $$
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