$\text{f}\left(\text{x}\right)=\{\begin{array}{l}\left[\mathrm{x}\right],\mathrm{x}\le 2\\ 0,\mathrm{x}>2\end{array}=\{\begin{array}{l}0,0\le \mathrm{x}<1\\ 1,1\le \mathrm{x}<2\\ 2,\mathrm{x}=2\\ 0,\mathrm{x}>2\end{array}$

$\therefore \text{\hspace{0.17em}f}\left({\text{x}}^{\text{2}}\right)=\{\begin{array}{l}0,0\le {\mathrm{x}}^2<1\\ 1,1\le {\mathrm{x}}^2<2\\ 2,{\mathrm{x}}^2=2\\ 0,{\mathrm{x}}^2>2\end{array}$

 $\text{f}(\text{x }+\text{ 1})=\{\begin{array}{l}0,0\le \mathrm{x}+1<1\\ 1,1\le \mathrm{x}+1<2\\ 2,\mathrm{x}+1=2\\ 0,\mathrm{x}+1>2\end{array}$

 $\mathrm{I}=\underset{-1}{\overset{2}{\int }}\frac{\mathrm{xf}\left({\mathrm{x}}^2\right)}{2+\mathrm{f}(\mathrm{x}+1)}\mathrm{dx}=\underset{-1}{\overset{1}{\int }}0\mathrm{dx}+\underset{1}{\overset{\sqrt{2}}{\int }}\frac{\mathrm{x}\text{\hspace{0.17em}}\mathrm{·}1}{2+0}\mathrm{dx}+\underset{\sqrt{2}}{\overset{2}{\int }}0\mathrm{dx}={\left(\frac{{\mathrm{x}}^2}{4}\right)}_1^{\sqrt{2}}=\left(\frac{2-1}{4}\right)=\frac{1}{4}.$

 $\therefore 4\mathrm{I}-1=0$