\mathrm{The}\mathrm{value}\mathrm{of}{\int }_0^3\frac{e^x+e^{-}}{\left[x\right]!}d\mathrm{x \; equals\; (Here\; [.]\; denotes\; the\; greatest\; integer\; function)}

To find the value of the given integral, we first state the correct answer clearly:
The correct option is:
$\frac{1}{2}(e^2+e^3-e^{-2}-e^{-3})$

Step-by-Step Explanation:

The given integral is:
$I={\int }_0^3\frac{e^x+e^{-x}}{\left[x\right]!}dx$
where $\left[x\right]$ represents the greatest integer function.

Since the value of $\left[x\right]$ changes at every integer value, we can partition the interval of integration $[0,3]$ into three sub-intervals: $[0,1)$, $[1,2)$, and $[2,3]$.

Let's determine the value of $\left[x\right]!$ in each of these intervals:
1. For $0\le x<1$, we have $\left[x\right]=0$, so $\left[x\right]!=0!=1$.
2. For $1\le x<2$, we have $\left[x\right]=1$, so $\left[x\right]!=1!=1$.
3. For $2\le x<3$, we have $\left[x\right]=2$, so $\left[x\right]!=2!=2$.

Now, we split the integral using these intervals:
$I={\int }_0^1\frac{e^x+e^{-x}}{1}dx+{\int }_1^2\frac{e^x+e^{-x}}{1}dx+{\int }_2^3\frac{e^x+e^{-x}}{2}dx$

Combining the first two integrals, we get:
$I={\int }_0^2(e^x+e^{-x})dx+\frac{1}{2}{\int }_2^3(e^x+e^{-x})dx$

Integrating the general term:
$\int (e^x+e^{-x})dx=e^x-e^{-x}$

Applying the integration limits to both parts:
$I={[e^x-e^{-x}]}_0^2+\frac{1}{2}{[e^x-e^{-x}]}_2^3$

Evaluating the limits:
$I=(e^2-e^{-2})-(e^0-e^0)+\frac{1}{2}(e^3-e^{-3}-e^2+e^{-2})$
Since $e^0-e^0=1-1=0$, we have:
$I=e^2-e^{-2}+\frac{1}{2}{e}^3-\frac{1}{2}{e}^{-3}-\frac{1}{2}{e}^2+\frac{1}{2}{e}^{-2}$

Grouping the terms together:
$I=(1-\frac{1}{2})e^2+\frac{1}{2}{e}^3+(\frac{1}{2}-1)e^{-2}-\frac{1}{2}{e}^{-3}$
$I=\frac{1}{2}{e}^2+\frac{1}{2}{e}^3-\frac{1}{2}{e}^{-2}-\frac{1}{2}{e}^{-3}$

Factoring out $\frac{1}{2}$:
$I=\frac{1}{2}(e^2+e^3-e^{-2}-e^{-3})$