Let α and β be real numbers such that

\underset{x\to 0}{lim}\frac{1}{x^3}[\alpha {\int }_1^x\frac{1}{1-t^2}dt+\beta x\cos x]=2

\mathrm{<br>}

Then the value of \alpha +\beta  is ______.

The correct answer is 2.40.

To find the value of $\alpha +\beta$, we analyze the given limit equation:
$\underset{x\to 0}{lim}\frac{1}{x^3}\left[\frac{\alpha }{2}{\int }_0^x\frac{1}{1-t^2}dt+\beta xcosx\right]=2$

Let us expand the terms inside the limit using Taylor series expansions around $x=0$.
First, we expand the integrand:
$\frac{1}{1-t^2}=1+t^2+t^4+...$

Integrating this term by term from $0$ to $x$ gives:
${\int }_0^x\frac{1}{1-t^2}dt={\left[t+\frac{t^3}{3}+...\right]}_0^x=x+\frac{x^3}{3}+O\left(x^5\right)$

Next, we expand the cosine function:
$cosx=1-\frac{x^2}{2}+O\left(x^4\right)$
Multiplying by $\beta x$ yields:
$\beta xcosx=\beta x-\frac{\beta x^3}{2}+O\left(x^5\right)$

Substitute these series expansions back into the limit expression:
$\underset{x\to 0}{lim}\frac{1}{x^3}\left[\frac{\alpha }{2}\left(x+\frac{x^3}{3}\right)+\beta x-\frac{\beta x^3}{2}+O\left(x^5\right)\right]=2$
Grouping the terms by powers of $x$:
$\underset{x\to 0}{lim}\frac{1}{x^3}\left[x\left(\frac{\alpha }{2}+\beta \right)+x^3\left(\frac{\alpha }{6}-\frac{\beta }{2}\right)+O\left(x^5\right)\right]=2$

For the limit to exist and equal the finite value of $2$, the coefficient of the $x$ term must be zero:
$\frac{\alpha }{2}+\beta =0\Rightarrow \alpha +2\beta =0$

When this condition is satisfied, the limit simplifies to the coefficient of the $x^3$ term:
$\frac{\alpha }{6}-\frac{\beta }{2}=2$

Now, we solve the system of linear equations. From the first equation, we get $\alpha =-2\beta$. Substituting this into the second equation:
$\frac{-2\beta }{6}-\frac{\beta }{2}=2$
Simplifying the expression:
$-\frac{\beta }{3}-\frac{\beta }{2}=2$
Finding a common denominator:
$-\frac{5\beta }{6}=2\Rightarrow \beta =-\frac{12}{5}=-2.40$

Substituting $\beta =-2.40$ back to find $\alpha$:
$\alpha =-2\beta =-2(-2.40)=4.80$

Thus, we compute the sum $\alpha +\beta$:
$\alpha +\beta =4.80+(-2.40)=2.40$