Evaluate the definite integral. \[ \int_{1}^{6} \frac{e^{1 / x}}{x^{2}} d x \]

To evaluate the definite integral \[ \int_{1}^{6} \frac{e^{1/x}}{x^2} \, dx, \] we can use a substitution. Let \[ u = \frac{1}{x} \implies du = -\frac{1}{x^2} \, dx \implies dx = -x^2 \, du = -\frac{1}{u^2} \, du, \] where we use the fact that \( x = \frac{1}{u} \). Next, we need to change the limits of integration. When \( x = 1 \), \( u = 1 \), and when \( x = 6 \), \( u = \frac{1}{6} \). Rewriting the integral in terms of \( u \): \[ \int_{1}^{6} \frac{e^{1/x}}{x^2} \, dx = \int_{1}^{1/6} e^u (-du) = \int_{1/6}^{1} e^u \, du. \] Now we can evaluate the integral: \[ \int e^u \, du = e^u + C. \] Thus, \[ \int_{1/6}^{1} e^u \, du = \left[ e^u \right]_{1/6}^{1} = e^{1} - e^{1/6}. \] Now we can express the final result: \[ \int_{1}^{6} \frac{e^{1/x}}{x^2} \, dx = e - e^{1/6}. \] So, the value of the definite integral is \[ \boxed{e - e^{1/6}}. \]