Answer
$$\int_{1}^{3}\frac{e^{\frac{3}{x}}}{x^{2}}dx=\frac{1}{3}(e^{3}-e)$$
Work Step by Step
Let:
$$t =\frac{3}{x},\,dx=-\frac{3}{t^{2}}dt$$
$$\int_{1}^{3}\frac{e^{\frac{3}{x}}}{x^{2}}dx=\int_{3}^{1}\frac{e^{t}}{\frac{9}{t^{2}}}(-\frac{3}{t^{2}})dt=-\frac{1}{3}\int_{3}^{1}e^{t}dt$$
$$=-\frac{1}{3}\left [ e^{t} \right ]_{3}^{1}=\frac{1}{3}(e^{3}-e)$$
