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: 68

Answer

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

Work Step by Step

\[\begin{gathered} \int_{}^{} {\frac{{{e^{2x}}}}{{{e^{2x}} + 1}}\,dx} \hfill \\ \hfill \\ u = {e^{2x}} + 1\,\,\,\,then\,\,\,du = 2{e^{2x}}dx \hfill \\ \hfill \\ apply\,\,the\,\,\,substitution \hfill \\ \hfill \\ = \frac{1}{2}\int_{}^{} {\frac{{du}}{u}} \hfill \\ \hfill \\ integrate \hfill \\ \hfill \\ = \frac{1}{2}\ln \left| u \right| + C \hfill \\ \hfill \\ replace\,\,u\,\,with\,\,\,u = {e^{2x}} + 1 \hfill \\ \hfill \\ = \frac{1}{2}\ln \,\left( {{e^{2x}} + 1} \right) + C \hfill \\ \hfill \\ \end{gathered} \]
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