Calculus: Early Transcendentals (2nd Edition)

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

Chapter 7 - Integration Techniques - 7.6 Other Integration Strategies - 7.6 Exercises: 38

Answer

\[ = \ln \,\,\left| {\frac{{\sqrt {1 + 4{e^t}} - 1}}{{\sqrt {1 + 4{e^t}} + 1}}} \right| + C\]

Work Step by Step

\[\begin{gathered} \int_{}^{} {\frac{{dt}}{{\sqrt {1 + 4{e^t}} }}} \hfill \\ \hfill \\ Rewrite\,\,\, \hfill \\ \hfill \\ = \frac{1}{4}\int_{}^{} {\frac{{4{e^t}dt}}{{{e^t}\sqrt {1 + 4{e^t}} }}} \hfill \\ \hfill \\ set\,\,u = 4{e^t}\,\,\,\,then\,\,\,\,du = 4{e^t}dt \hfill \\ \hfill \\ = \int_{}^{} {\frac{{du}}{{u\sqrt {1 + u} }}} \hfill \\ \hfill \\ integrate\,\,\,by\,\,tables \hfill \\ \hfill \\ = \frac{1}{{\sqrt 1 }}\ln \,\,\left| {\frac{{\sqrt {1 + u} - 1}}{{\sqrt {1 + u} + 1}}} \right| + C \hfill \\ \hfill \\ substitute\,\,back\,\,u = 4{e^t}\,\,and\,\,simplify \hfill \\ \hfill \\ = \ln \,\,\left| {\frac{{\sqrt {1 + 4{e^t}} - 1}}{{\sqrt {1 + 4{e^t}} + 1}}} \right| + C \hfill \\ \end{gathered} \]
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