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: 26

Answer

$$\frac{7}{{24}}$$

Work Step by Step

$$\eqalign{ & \int_{\ln 2}^{\ln 3} {\frac{{{e^x} + {e^{ - x}}}}{{{e^{2x}} - 2 + {e^{ - 2x}}}}} \cr & {\text{factoring the denominator}} \cr & \int_{\ln 2}^{\ln 3} {\frac{{{e^x} + {e^{ - x}}}}{{{{\left( {{e^x} - {e^{ - x}}} \right)}^2}}}} dx \cr & {\text{substitute }}u = {e^x} - {e^{ - x}},{\text{ }}du = \left( {{e^x} + {e^{ - x}}} \right)dx \cr & {\text{express the limits in terms of }}u \cr & x = \ln 2{\text{ implies }}u = {e^{\ln 2}} - {e^{ - \ln 2}} = 3/2 \cr & x = \ln 3{\text{ implies }}u = {e^{\ln 3}} - {e^{ - \ln 3}} = 8/3 \cr & {\text{the entire integration is carried out as follows}} \cr & \int_{\ln 2}^{\ln 3} {\frac{{{e^x} + {e^{ - x}}}}{{{{\left( {{e^x} - {e^{ - x}}} \right)}^2}}}} dx = \int_{3/2}^{8/3} {\frac{1}{{{u^2}}}} du \cr & {\text{find the antiderivative}} \cr & = \left. { - \frac{1}{u}} \right|_{3/2}^{8/3} \cr & {\text{use the fundamental theorem}} \cr & = - \left( {\frac{3}{8} - \frac{2}{3}} \right) \cr & {\text{simplify}} \cr & = \frac{7}{{24}} \cr} $$
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