If the value of the integral ${\int }_0^5\frac{x+\left[x\right]}{e^{x-\left[x\right]}}\mathrm{d}x=\alpha e^{-1}+\beta ,$ where $\alpha , \beta \in R, 5\alpha +6\beta =0,$ and $\left[x\right]$ denotes the greatest integer less than or equal to $x;$ then the value of $(\alpha +\beta {)}^2$ is equal to :

${\int }_0^5\frac{x+\left[x\right]}{e^{x-\left[x\right]}}dx=\alpha e^{-1}+\beta$

$\Rightarrow {\int }_0^1\frac{x}{e^x}dx+{\int }_1^2\frac{x+1}{e^{x-1}}dx+{\int }_2^3\frac{x+2}{e^{x-2}}dx+{\int }_3^4\frac{x+3}{e^{x-3}}dx+{\int }_4^5\frac{x+4}{e^{x-4}}dx$

Let $x-1=p, x-2=q, x-3=r, x-4=w$

$\Rightarrow {\int }_0^1\frac{x}{e^x}dx+{\int }_0^1\frac{p+2}{e^p}dp+{\int }_0^1\frac{q+4}{e^q}dq+{\int }_0^1\frac{r+6}{e^r}dr+{\int }_0^1\frac{w+8}{e^w}dw$

$\Rightarrow {\int }_0^1\frac{x}{e^x}dx+{\int }_0^1\frac{x+2}{e^x}dx+{\int }_0^1\frac{x+4}{e^x}dx+{\int }_0^1\frac{x+6}{e^x}dx+{\int }_0^1\frac{x+8}{e^x}dx$

$={\int }_0^1\frac{5x+20}{e^x}dx=5{\int }_0^1\frac{x+4}{e^x}dx$

$=5{\left[\left(x+4\right)\frac{{\mathrm{e}}^{-x}}{-1}\right]}_0^1-{\int }_0^1\frac{e^{-x}}{-1}dx$

$=5\left[-5e^{-1}+4+\left(-e^{-1}+1\right)\right]$$=5\left(-6e^{-1}+5\right)=-30e^{-1}+25$

$\alpha =-30,\beta =25$

$(\alpha +\beta {)}^2=25$