Calculus: Early Transcendentals (2nd Edition)

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

Chapter 7 - Integration Techniques - 7.5 Partial Fractions - 7.5 Exercises: 73

Answer

\[ = - \frac{1}{{2\,\left( {1 + {e^{2x}}} \right)}} + C\]

Work Step by Step

\[\begin{gathered} \int_{}^{} {\frac{{dx}}{{\,{{\left( {{e^x} + {e^{ - x}}} \right)}^2}}}} \hfill \\ \hfill \\ rewrite\,\, \hfill \\ \hfill \\ \int_{}^{} {\frac{{dx}}{{\,{{\left( {{e^x} + {e^{ - x}}} \right)}^2}}}} = \int_{}^{} {\frac{{{e^{2x}}}}{{\,{{\left( {1 + {e^{2x}}} \right)}^2}}}dx} \hfill \\ \hfill \\ Let\,,\,\,1 + {e^{2x}} = t\,\,\,\,\,\,then\,\,\,\,\,\,2{e^{2x}}dx = dt \hfill \\ \hfill \\ apply\,\,the\,\,substitution \hfill \\ \hfill \\ \int_{}^{} {\frac{{dx}}{{\,{{\left( {{e^x} + {e^{ - x}}} \right)}^2}}}} = \frac{1}{2}\int_{}^{} {\frac{{dt}}{{{t^2}}}} \hfill \\ \hfill \\ {\text{integrate}} \hfill \\ \hfill \\ = - \frac{1}{{2t}} + C \hfill \\ \hfill \\ substitute\,\,for\,\,\,\left( t \right) \hfill \\ \hfill \\ = - \frac{1}{{2\,\left( {1 + {e^{2x}}} \right)}} + C \hfill \\ \end{gathered} \]
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