What is the integral $\int\limits_1^3\frac{e^{\frac{1}{x}}}{x^{2}}dx$? What is

$$\int_1^3 \frac{e^{\frac{1}{x}}}{x^{2}} dx ?$$

How do you approach solving this?
Let $u = \frac{1}{x}$. Then $du = -\frac{1}{x^2} \ dx$. Hence we now have

$$- \int_{x=1}^{x=3} e^u \ du = -e^u = -e^\frac{1}{x}$$

Plug in the bounds now to get your answer of $e-e^\frac{1}{3}$.
The following integral can be computed by substitution, $t=\frac{1}{x}$ 
$$\int_1^3 \frac{e^{\frac{1}{x}}}{x^2}dx=-\int_1^\frac{1}{3} e^tdt=e-e^\frac{1}{3}$$