Calculus: Early Transcendentals (2nd Edition)

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

Chapter 7 - Integration Techniques - 7.1 Basic Approaches - 7.1 Exercises: 13

Answer

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

Work Step by Step

\[\begin{gathered} = \int_{}^{} {\frac{{{e^x}}}{{{e^x} + 1}}dx} \hfill \\ \hfill \\ set\,\,{e^x} + 1 = u\,\,then\,\,\,{e^x}dx = du \hfill \\ \hfill \\ integral\,\,becomes \hfill \\ \hfill \\ = \int_{}^{} {\frac{{{e^x}}}{{{e^x} + 1}}dx} = \int_{}^{} {\frac{{du}}{u}} \hfill \\ \hfill \\ integrate \hfill \\ \hfill \\ = \ln u + C \hfill \\ \hfill \\ substituting\,back\,\,\,u = {e^x} + 1 \hfill \\ \hfill \\ = \ln \,\left( {{e^x} + 1} \right) + C \hfill \\ \end{gathered} \]
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