1. We need to evaluate $I=\int_1^3 [x^2-2x-2] \, dx,$ where $[\,\cdot\,]$ denotes the greatest integer function.

2. First simplify the expression inside: $x^2-2x-2=(x-1)^2-3.$ Let $f(x)=x^2-2x-2.$ On $[1,3]$, this is increasing for $x>1$ because $f'(x)=2x-2\ge 0.$ Also, $f(1)=-3, \qquad f(3)=1.$ So $f(x)$ moves from $-3$ to $1$ on $[1,3]$.

3. Since we need $[f(x)]$, find where $f(x)$ crosses integers:

• For $f(x)=-2$: $x^2-2x-2=-2 \implies x^2-2x=0 \implies x(x-2)=0.$ In $[1,3]$, this gives $x=2.$

• For $f(x)=-1$: $x^2-2x-2=-1 \implies x^2-2x-1=0.$ So $x=1\pm \sqrt2.$ In $[1,3]$, this gives $x=1+\sqrt2.$

• For $f(x)=0$: $x^2-2x-2=0 \implies x=1\pm \sqrt3.$ In $[1,3]$, this gives $x=1+\sqrt3.$

4. Therefore, the floor value changes as follows:

• For $1\le x<2$, we have $-3\le f(x)<-2 \implies [f(x)]=-3.$

• For $2\le x<1+\sqrt2$, we have $-2\le f(x)<-1 \implies [f(x)]=-2.$

• For $1+\sqrt2\le x<1+\sqrt3$, we have $-1\le f(x)<0 \implies [f(x)]=-1.$

• For $1+\sqrt3\le x<3$, we have $0\le f(x)<1 \implies [f(x)]=0.$ At $x=3$, $f(3)=1$, but a single point does not affect the integral.

5. Now integrate piecewise: $I=\int_1^2 (-3)\,dx+\int_2^{1+\sqrt2} (-2)\,dx+\int_{1+\sqrt2}^{1+\sqrt3} (-1)\,dx+\int_{1+\sqrt3}^3 0\,dx.$

Compute each part: $\int_1^2 (-3)\,dx=-3(2-1)=-3,$ $\int_2^{1+\sqrt2} (-2)\,dx=-2\big((1+\sqrt2)-2\big)=-2(\sqrt2-1)=-2\sqrt2+2,$ $\int_{1+\sqrt2}^{1+\sqrt3} (-1)\,dx=-\big((1+\sqrt3)-(1+\sqrt2)\big)=-(\sqrt3-\sqrt2)=-\sqrt3+\sqrt2.$

So, $I=-3+(-2\sqrt2+2)+(-\sqrt3+\sqrt2).$ Simplify: $I=-1-\sqrt2-\sqrt3.$

6. Hence, $\boxed{I=-\sqrt2-\sqrt3-1}.$

7. Comparing with the options:

• Option A: $-5$
• Option B: $-\sqrt2-\sqrt3+1$
• Option C: $-4$
• Option D: $-\sqrt2-\sqrt3-1$

So the correct option is $\boxed{\text{D}}.$