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: 51

Answer

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

Work Step by Step

\[\begin{gathered} \int_{}^{} {\frac{{{e^x}}}{{{e^{2x}} + 2{e^x} + 1}}\,dx} \hfill \\ \hfill \\ factoring \hfill \\ \hfill \\ = \int_{}^{} {\frac{{{e^x}}}{{\,{{\left( {{e^x} + 1} \right)}^2}}}\,dx} \hfill \\ \hfill \\ set\,\,{e^x} + 1 = t\,\,\,\, \to \,\,\,\,{e^x}dx = dt \hfill \\ \hfill \\ integral\,becomes \hfill \\ \hfill \\ \int_{}^{} {\frac{{{e^x}}}{{{e^{2x}} + 2{e^x} + 1}}\,dx} = \int_{}^{} {\frac{{dt}}{{{t^2}}}} \hfill \\ \hfill \\ \int_{}^{} {{t^{ - 2}}dt} \hfill \\ \hfill \\ {\text{use }}\int {{x^n}dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C} \hfill \\ \hfill \\ - \frac{1}{t} + C \hfill \\ \hfill \\ substituting\,back\,\,t = {e^x} + 1 \hfill \\ \hfill \\ = \frac{{ - 1}}{{\,\left( {{e^x} + 1} \right)}} + C \hfill \\ \end{gathered} \]
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