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