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: 31

Answer

\[ - 3\ln \left| x \right| - 10{e^{ - 0.4}} + {e^{0.1}}x + C\]

Work Step by Step

\[\begin{gathered} \int_{}^{} {\,\left( { - \frac{3}{x} + 4{e^{ - 0.4x}} + {e^{0.1}}} \right)dx} \hfill \\ By\,\,the\,\,sum\,\,and\,\,differentiate\,\,rule \hfill \\ - \int_{}^{} {\frac{3}{x}dx} + \int_{}^{} {4{e^{ - 0.4x}}dx} + \int_{}^{} {{e^{0.1}}\,dx} \hfill \\ Use\,\,indefinite\,\,integrals \hfill \\ \int_{}^{} {{x^{ - 1}}dx} = \ln \left| x \right| + C\,\,,\,\,\int_{}^{} {{e^{kx}}dx} = \frac{{{e^{kx}}}}{k} + C \hfill \\ Then \hfill \\ - 3\ln \left| x \right| + 4\,\left( {\frac{{{e^{ - 0.4x}}}}{{ - 0.4}}} \right) + {e^{0.1}}x + C \hfill \\ Simplifying \hfill \\ - 3\ln \left| x \right| + 4\,\left( { - \frac{5}{2}{e^{ - 0.4x}}} \right) + {e^{0.1}}x + C \hfill \\ - 3\ln \left| x \right| - 10{e^{ - 0.4}} + {e^{0.1}}x + C \hfill \\ \hfill \\ \end{gathered} \]
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