Let $$\alpha$$ and $$\beta$$ be the real numbers such that

$$\lim_{x \to 0} \frac{1}{x^3} \left( \frac{\alpha}{2} \int_0^x \frac{1}{1 - t^2}\, dt + \beta x \cos x \right) = 2.$$

Then the value of $$\alpha + \beta$$ is ________.

Let $$F(x)=\dfrac{1}{x^{3}}\left(\dfrac{\alpha}{2}\int_{0}^{x}\dfrac{1}{1-t^{2}}\;dt+\beta x\cos x\right)$$ and we want $$\displaystyle\lim_{x\to 0}F(x)=2$$.

Step 1 : Series expansion of the integral
Use the geometric series for small $$|t|$$: $$\dfrac{1}{1-t^{2}}=1+t^{2}+t^{4}+t^{6}+\dots$$

Integrate term-by-term from $$0$$ to $$x$$:
$$\int_{0}^{x}\dfrac{1}{1-t^{2}}\;dt=\int_{0}^{x}\!\!(1+t^{2}+t^{4}+\dots)\;dt = x+\dfrac{x^{3}}{3}+\dfrac{x^{5}}{5}+O(x^{7}).$$

Step 2 : Series expansion of $$x\cos x$$
The Maclaurin series of $$\cos x$$ is $$1-\dfrac{x^{2}}{2}+\dfrac{x^{4}}{24}+O(x^{6})$$, hence
$$x\cos x=x-\dfrac{x^{3}}{2}+\dfrac{x^{5}}{24}+O(x^{7}).$$

Step 3 : Numerator as a power series
Write the numerator $$N(x)$$ up to order $$x^{3}$$:
$$N(x)=\dfrac{\alpha}{2}\!\left(x+\dfrac{x^{3}}{3}+O(x^{5})\right) +\beta\!\left(x-\dfrac{x^{3}}{2}+O(x^{5})\right).$$

Collect coefficients:

• Coefficient of $$x$$: $$C_{1}= \dfrac{\alpha}{2}+\beta.$$
• Coefficient of $$x^{3}$$: $$C_{3}= \dfrac{\alpha}{2}\cdot\dfrac{1}{3}+\beta\!\left(-\dfrac{1}{2}\right)=\dfrac{\alpha}{6}-\dfrac{\beta}{2}.$$

Step 4 : Finite limit condition
Since we divide by $$x^{3}$$, the term in $$x$$ must vanish for the limit to remain finite:
$$C_{1}=0 \;\;\Longrightarrow\;\; \dfrac{\alpha}{2}+\beta=0\;\;\Longrightarrow\;\;\beta=-\dfrac{\alpha}{2}.$$

With this relation, the numerator behaves as
$$N(x)=\Bigl(\dfrac{\alpha}{6}-\dfrac{\beta}{2}\Bigr)x^{3}+O(x^{5}).$$

Substitute $$\beta=-\dfrac{\alpha}{2}$$:
$$C_{3}= \dfrac{\alpha}{6}-\Bigl(-\dfrac{\alpha}{4}\Bigr)=\dfrac{\alpha}{6}+\dfrac{\alpha}{4}=\dfrac{5\alpha}{12}.$$

Step 5 : Impose the required limit
Thus,
$$F(x)=\dfrac{N(x)}{x^{3}}=\dfrac{5\alpha}{12}+O(x^{2}).$$
Taking $$x\to 0$$ gives $$\displaystyle\lim_{x\to 0}F(x)=\dfrac{5\alpha}{12}.$$

We need this limit to equal $$2$$, so
$$\dfrac{5\alpha}{12}=2\;\;\Longrightarrow\;\;\alpha=\dfrac{24}{5}=4.8.$$

Step 6 : Find $$\beta$$ and $$\alpha+\beta$$
$$\beta=-\dfrac{\alpha}{2}=-\dfrac{4.8}{2}=-2.4.$$
Therefore,
$$\alpha+\beta=4.8-2.4=2.4.$$

The required value is $$\boxed{2.4}$$, which lies in the interval 2.35 - 2.45.