Let $$[\cdot]$$ denote the greatest integer function. Then the value of $$\int_0^3 \left(\frac{e^x + e^{-x}}{[x]!}\right) dx$$ is :

Over each sub-interval in which $$[x]$$ is constant, the integrand becomes a simple exponential quotient.
Since the limits are 0 to 3, the greatest integer function $$[x]$$ takes the values 0, 1 and 2 only.

Case 1: $$0 \le x \lt 1$$  ⟹  $$[x]=0$$, so $$[x]!=0! = 1$$.
Integral on this interval:

$$I_1 = \int_0^{1} \left(e^{x}+e^{-x}\right)\,dx$$

Using $$\int e^{x}\,dx = e^{x}$$ and $$\int e^{-x}\,dx = -e^{-x}$$:

$$I_1 = \bigl[e^{x}\bigr]_0^{1} + \bigl[-e^{-x}\bigr]_0^{1} = (e-1) + ( -e^{-1} + 1 ) = e-\frac1e.$$

Case 2: $$1 \le x \lt 2$$  ⟹  $$[x]=1$$, so $$[x]!=1! = 1$$.
Integral on this interval:

$$I_2 = \int_{1}^{2} \left(e^{x}+e^{-x}\right)\,dx = (e^{2}-e) + ( -e^{-2}+e^{-1} ) = e^{2} + \frac1e - e - \frac1{e^{2}}.$$

Case 3: $$2 \le x \lt 3$$  ⟹  $$[x]=2$$, so $$[x]!=2! = 2$$.
Integral on this interval:

$$I_3 = \int_{2}^{3} \frac{e^{x}+e^{-x}}{2}\,dx = \frac12 \left[ \int_{2}^{3} e^{x}\,dx + \int_{2}^{3} e^{-x}\,dx \right].$$

Compute the two integrals separately:
$$\int_{2}^{3} e^{x}\,dx = e^{3}-e^{2}, \quad \int_{2}^{3} e^{-x}\,dx = \bigl[-e^{-x}\bigr]_{2}^{3} = -e^{-3}+e^{-2}.$$

Hence

$$I_3 = \frac12\left(e^{3}-e^{2}-\frac1{e^{3}}+\frac1{e^{2}}\right).$$

Add the three parts:

$$I = I_1 + I_2 + I_3$$ $$\;\; = \left(e-\frac1e\right) + \left(e^{2}+\frac1e-e-\frac1{e^{2}}\right) + \frac12\left(e^{3}-e^{2}-\frac1{e^{3}}+\frac1{e^{2}}\right).$$

Simplify term-by-term:

• The terms $$e$$ and $$-e$$ cancel.
• The terms $$\tfrac1e$$ and $$-\tfrac1e$$ cancel.

Remaining terms:

$$I = e^{2}-\frac1{e^{2}} + \frac12\left(e^{3}-e^{2}-\frac1{e^{3}}+\frac1{e^{2}}\right)$$ $$\;\; = \left(e^{2}-\frac1{e^{2}}\right) + \left(\frac12e^{3}-\frac12e^{2}-\frac1{2e^{3}}+\frac1{2e^{2}}\right)$$ $$\;\; = \frac12e^{3} + \frac12e^{2} - \frac1{2e^{2}} - \frac1{2e^{3}}.$$(Combine like terms.)

Factor out $$\frac12$$:

$$I = \frac12\left(e^{3}+e^{2}-\frac1{e^{2}}-\frac1{e^{3}}\right).$$

This matches Option B.

Hence the required value is
$$\boxed{\displaystyle \frac12\left(e^{2}+e^{3}-\frac1{e^{2}}-\frac1{e^{3}}\right)}.$$