Answer
$e-\sqrt{e}$
Work Step by Step
$\int_{1}^{2}\frac{e^{1/x}}{x^2}~dx$
Let $u = \frac{1}{x}$
$\frac{du}{dx} = -\frac{1}{x^2}$
$dx = -x^2~du$
When $x = 1$, then $u = 1$
When $x = 2$, then $u = \frac{1}{2}$
$\int_{1}^{1/2} \frac{e^u}{x^2}~(-x^2~du)$
$= \int_{1}^{1/2}-e^u~du$
$= \int_{1/2}^{1}e^u~du$
$= e^u~\vert_{1/2}^{1}$
$= e^1-e^{1/2}$
$= e-\sqrt{e}$