Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 7 - Integration - 7.1 Antiderivatives - 7.1 Exercises: 32

Answer

\[9\ln \left| x \right| + \frac{{15}}{2}{e^{ - 0.4x}} + C\]

Work Step by Step

\[\begin{gathered} \int_{}^{} {\,\left( {\frac{9}{x} - 3{e^{ - 0.4x}}} \right)dx} \hfill \\ By\,\,the\,\,sum\,\,and\,\,differentiate\,\,rule \hfill \\ \int_{}^{} {\frac{9}{x}dx - \int_{}^{} {3{e^{ - 0.4x}}dx} } \hfill \\ 9\int_{}^{} {{x^{ - 1}}dx} - 3\int_{}^{} {{e^{ - 0.4x}}dx} \hfill \\ Use\,\,indefinite\,\,integrals \hfill \\ \int_{}^{} {{x^{ - 1}}dx = \ln \left| x \right| + C\,\,,\,\,\,\int_{}^{} {{e^{kx}}dx} = \frac{{{e^{kx}}}}{k} + C} \hfill \\ 9\ln \left| x \right| - \frac{{3{e^{ - 0.4x}}}}{{ - 0.4}} + C \hfill \\ Simplifying \hfill \\ 9\ln \left| x \right| + \frac{{15}}{2}{e^{ - 0.4x}} + C \hfill \\ \end{gathered} \]
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