Answer
$\displaystyle\int_1^3{\frac{e^{\frac{3}{x}}}{x^2}dx}=\frac{1}{3}\left( e^3-e \right)$
Work Step by Step
$I=\displaystyle\int_1^3{\frac{e^{\frac{3}{x}}}{x^2}dx}\\
I=\displaystyle\int_1^3{\frac{e^{3x^{-1}}}{x^2}dx}\\
I=\displaystyle\int_1^3{e^{3x^{-1}}x^{-2}\ dx}$
$\displaystyle{\left[\begin{matrix}
u=x^{-1}& dx=-\frac{du}{x^{-2}}\\
\end{matrix}\right]}
$
$I=\displaystyle\int_1^{\frac{1}{3}}{e^{3u}x^{-2}.-\frac{du}{x^{-2}}}\\
I=\displaystyle-\int_1^{\frac{1}{3}}{e^{3u}du}\\
I=\displaystyle-\left[ \frac{1}{3}e^{3u} \right] _{1}^{\frac{1}{3}}\\
I=\displaystyle\frac{1}{3}\left( e^3-e \right)\\
$
