Answer
$e-\sqrt{e}$
Work Step by Step
Let $u=\frac{1}{x}$. Then $u=x^{-1}$, and $du=-1x^{-2}=-\frac{1}{x^2}$, so $-du=\frac{1}{x^2}$.
$\int_1^2\frac{e^{1/x}}{x^2}dx$
$=\int_{\frac{1}{1}}^{\frac{1}{2}}e^u*(-1) du$
$=-\int_1^{\frac{1}{2}}e^udu$
$=-e^u|_1^\frac{1}{2}$
$=-(e^\frac{1}{2}-e^1)$
$=-(\sqrt{e}-e)$
$=e-\sqrt{e}$