Expand each piece as a Maclaurin series.

For the integral, $\frac{1}{1-t^2}=1+t^2+t^4+\cdots$, so

${\int }_0^x\frac{dt}{1-t^2}=x+\frac{x^3}{3}+\frac{x^5}{5}+\cdots .$

And $\cos x=1-\frac{x^2}{2}+\frac{x^4}{24}-\cdots$, so $x\cos x=x-\frac{x^3}{2}+\frac{x^5}{24}-\cdots .$

Therefore the bracketed expression equals

$\frac{\alpha }{2}\left(x+\frac{x^3}{3}+\cdots \right)+\beta \left(x-\frac{x^3}{2}+\cdots \right)=\left(\frac{\alpha }{2}+\beta \right)x+\left(\frac{\alpha }{6}-\frac{\beta }{2}\right)x^3+O(x^5).$

Dividing by $x^3$ and requiring a finite limit equal to $2$ forces the $x$-coefficient to vanish and the $x^3$-coefficient to equal $2$:

$\frac{\alpha }{2}+\beta =0,\frac{\alpha }{6}-\frac{\beta }{2}=2.$

From the first equation $\alpha =-2\beta$. Substituting: $-\frac{\beta }{3}-\frac{\beta }{2}=2\Rightarrow -\frac{5\beta }{6}=2\Rightarrow \beta =-\frac{12}{5}$ and $\alpha =\frac{24}{5}.$

Hence $\alpha +\beta =\frac{24-12}{5}=\frac{12}{5}=\mathrm{2.40.}$

$\therefore$ The answer is $\boxed{2.40}$.