Evaluation of Definite Integral Involving Greatest Integer Function

To evaluate the definite integral $I = \int_{0}^{3} \left( \frac{e^x + e^{-x}}{[x]!} \right) dx$ where $[x]$ denotes the greatest integer function, we break the interval of integration $[0, 3]$ into sub-intervals based on the step values of $[x]$:

1. For $x \in [0, 1)$, $[x] = 0 \implies [x]! = 0! = 1$
2. For $x \in [1, 2)$, $[x] = 1 \implies [x]! = 1! = 1$
3. For $x \in [2, 3)$, $[x] = 2 \implies [x]! = 2! = 2$

Using these values, the integral $I$ can be split into three separate integrals: $I = \int_{0}^{1} \frac{e^x + e^{-x}}{0!} dx + \int_{1}^{2} \frac{e^x + e^{-x}}{1!} dx + \int_{2}^{3} \frac{e^x + e^{-x}}{2!} dx$

Substitute the factorial values into the integral: $I = \int_{0}^{1} (e^x + e^{-x}) dx + \int_{1}^{2} (e^x + e^{-x}) dx + \frac{1}{2} \int_{2}^{3} (e^x + e^{-x}) dx$

The first two integrals can be combined into a single integral from $0$ to $2$: $\int_{0}^{2} (e^x + e^{-x}) dx = \left[ e^x - e^{-x} \right]_{0}^{2} = \left( e^2 - e^{-2} \right) - \left( e^0 - e^0 \right) = e^2 - e^{-2}$

Now, evaluate the third integral: $\frac{1}{2} \int_{2}^{3} (e^x + e^{-x}) dx = \frac{1}{2} \left[ e^x - e^{-x} \right]_{2}^{3} = \frac{1}{2} \left( e^3 - e^{-3} - (e^2 - e^{-2}) \right)$

Adding these two results together gives: $I = (e^2 - e^{-2}) + \frac{1}{2} (e^3 - e^{-3}) - \frac{1}{2} (e^2 - e^{-2})$

Combine like terms: $I = \frac{1}{2} (e^2 - e^{-2}) + \frac{1}{2} (e^3 - e^{-3})$ $I = \frac{1}{2} \left( e^2 + e^3 - \frac{1}{e^2} - \frac{1}{e^3} \right)$

Hence, the correct option is B.