Let \(\alpha\) and \(\beta\) be the real numbers such that \(\lim _{x \rightarrow 0} \frac{1}{x^3}\left(\frac{\alpha}{2} \int_0^x \frac{1}{1-t^2} d t+\beta x \cos x\right)=2\). Then the value of \(\alpha+\beta\) is _______

\(\lim _{x \rightarrow 0} \frac{\frac{\alpha}{2} \int_0^x \frac{1}{1-t^2} d t+\beta x \cos x}{x^3}\)
\(\begin{aligned} & =\lim _{x \rightarrow 0} \frac{\frac{\alpha}{2}\left(\frac{1}{1-x^2}\right)+\beta \cos x-\beta x \sin x}{3 x^2} \\ & =\frac{\frac{\alpha}{2}\left(1-x^2\right)^{-1}+\beta\left(1-\frac{x^2}{2!}+\frac{x^4}{4!} \ldots\right)-\beta x\left(x-\frac{x^3}{3!}+\frac{x^5}{5!} \ldots\right)}{3 x^2} \\ & =\frac{\frac{\alpha}{2}\left(1+x^2+x^4 \ldots\right)+\beta\left(1-\frac{x^2}{2!}+\frac{x^4}{4!} \ldots\right)-\beta\left(x^2-\frac{x^4}{3!} \ldots\right)}{3 x^2}\end{aligned}\)
\(\begin{aligned} & =\frac{\left(\frac{\alpha}{2}+\beta\right)+\mathrm{x}^2\left(\frac{\alpha}{2}-\frac{\beta}{2}-\beta\right)+\mathrm{x}^4() \ldots}{3 \mathrm{x}^2}=2(\text { Given }) \\ & \therefore \frac{\alpha}{2}+\beta=0 \text { and } \frac{\alpha-3 \beta}{6}=2 \\ & \Rightarrow \alpha=-2 \beta \text { and } \alpha=12+3 \beta \\ & \Rightarrow \beta=-\frac{12}{5} \text { and } \alpha=\frac{24}{5}\end{aligned}\)
\(\therefore \alpha+\beta=\frac{12}{5}=2.40\)