Answer
$3e^{-1/x}+C$
Work Step by Step
$I=\displaystyle \int\frac{3e^{-1/x}}{x^{2}}dx=\int(3e^{-1/x})(\frac{dx}{x^{2}})$
Substitute: $\left[\begin{array}{llll}
u & =-\frac{1}{x}=-x^{-1} & & \\
& & & \\
du & =x^{-2}dx & \Rightarrow & \dfrac{dx}{x^{2}}=du
\end{array}\right]$
$I=\displaystyle \int 3e^{u}du$
$=3e^{u}+C$
bring back the variable $x$
= $3e^{-1/x}+C$
