Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.5 Exercises - Page 549: 32

Answer

$ - \frac{1}{4}\ln \left( {{e^{2x}} + 1} \right) - \frac{1}{2}{\tan ^{ - 1}}\left( {{e^x}} \right) + \frac{1}{2}\ln \left| {{e^x} - 1} \right| + C$

Work Step by Step

$$\eqalign{ & \int {\frac{{{e^x}}}{{\left( {{e^{2x}} + 1} \right)\left( {{e^x} - 1} \right)}}} dx \cr & {\text{Let }}t = {e^x},{\text{ }}dt = {e^x}dx \cr & {\text{Substituting}} \cr & \int {\frac{{{e^x}}}{{\left( {{e^{2x}} + 1} \right)\left( {{e^x} - 1} \right)}}} dx = \int {\frac{1}{{\left( {{t^2} + 1} \right)\left( {t - 1} \right)}}} dt \cr & {\text{Decompose into partial fractions}} \cr & \frac{1}{{\left( {{t^2} + 1} \right)\left( {{t^2} - 1} \right)}} = \frac{{At + B}}{{{t^2} + 1}} + \frac{C}{{t - 1}} \cr & 1 = \left( {At + B} \right)\left( {t - 1} \right) + C\left( {{t^2} + 1} \right) \cr & {\text{Distribute}} \cr & 1 = A{t^2} - At + Bt - B + C{t^2} + C \cr & {\text{Group terms}} \cr & 1 = \left( {A{t^2} + C{t^2}} \right) + \left( { - At + Bt} \right) + \left( { - B + C} \right) \cr & A + C = 0{\text{ }}\left( {\bf{1}} \right) \cr & - A + B = 0{\text{ }}\left( {\bf{2}} \right) \cr & - B + C = 1{\text{ }}\left( {\bf{3}} \right) \cr & {\text{Solving the system of linear equations we obtain}} \cr & A = - \frac{1}{2},{\text{ }}B = - \frac{1}{2},{\text{ }}C = \frac{1}{2} \cr & \frac{1}{{\left( {{t^2} + 1} \right)\left( {{t^2} - 1} \right)}} = \frac{{At + B}}{{{t^2} + 1}} + \frac{C}{{t - 1}} \cr & \frac{1}{{\left( {{t^2} + 1} \right)\left( {{t^2} - 1} \right)}} = \frac{{ - \frac{1}{2}t - \frac{1}{2}}}{{{t^2} + 1}} + \frac{{\frac{1}{2}}}{{t - 1}} \cr & \int {\frac{1}{{\left( {{t^2} + 1} \right)\left( {t - 1} \right)}}} dt = \int {\left( {\frac{{ - \frac{1}{2}t - \frac{1}{2}}}{{{t^2} + 1}} + \frac{{\frac{1}{2}}}{{t - 1}}} \right)} dt \cr & = - \frac{1}{2}\int {\frac{t}{{{t^2} + 1}}} dt - \frac{1}{2}\int {\frac{1}{{{t^2} + 1}}} dt + \frac{1}{2}\int {\frac{1}{{t - 1}}} dt \cr & = - \frac{1}{4}\int {\frac{{2t}}{{{t^2} + 1}}} dt - \frac{1}{2}\int {\frac{1}{{{t^2} + 1}}} dt + \frac{1}{2}\int {\frac{1}{{t - 1}}} dt \cr & {\text{Integrate}} \cr & {\text{ = }} - \frac{1}{4}\ln \left( {{t^2} + 1} \right) - \frac{1}{2}{\tan ^{ - 1}}t + \frac{1}{2}\ln \left| {t - 1} \right| + C \cr & {\text{Write in terms of }}x,{\text{ substitute }}{e^x}{\text{ for }}t \cr & {\text{ = }} - \frac{1}{4}\ln \left( {{e^{2x}} + 1} \right) - \frac{1}{2}{\tan ^{ - 1}}\left( {{e^x}} \right) + \frac{1}{2}\ln \left| {{e^x} - 1} \right| + C \cr} $$
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