Calculus with Applications (10th Edition)

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

Chapter 7 - Integration - 7.2 Substitution - 7.2 Exercises: 15

Answer

\[ = - {e^{\frac{1}{z}}} + C\]

Work Step by Step

\[\begin{gathered} \int_{}^{} {\frac{{{e^{2/z}}}}{{{z^2}}}dz} \hfill \\ Write\,\,the\,\,integrand\,\,as \hfill \\ \int_{}^{} {{e^{z - 1}}{z^{ - 2}}dz} \hfill \\ Let\,\,u = {z^{ - 1}} \hfill \\ So\,\,that\,\,du = - {z^{ - 2}}dz \hfill \\ Then \hfill \\ - \int_{}^{} {{e^{z - 1}}\,\left( { - {z^{ - 2}}} \right)dz} \hfill \\ - \int_{}^{} {{e^u}du} = - {e^u} + C \hfill \\ Substituting\,\,u = {z^{ - 1}}\,gives \hfill \\ = - {e^{{z^{ - 1}}}} + C \hfill \\ = - {e^{\frac{1}{z}}} + C \hfill \\ \hfill \\ \end{gathered} \]
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