[THOUGHT] I need to find the values of $a$ and $b$ such that the given limit equals 2. Let me first expand the integrand and the cosine function using Taylor series around $x = 0$.

The integral $\int_0^x \frac{1}{1-t^2} dt$ can be computed. We know that $\frac{1}{1-t^2} = \frac{1}{(1-t)(1+t)} = \frac{1}{2}\left(\frac{1}{1-t} + \frac{1}{1+t}\right)$.

So $\int_0^x \frac{1}{1-t^2} dt = \frac{1}{2}\left(-\ln(1-t) + \ln(1+t)\right)\Big|_0^x = \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)$.

Now I need to expand this as a Taylor series around $x = 0$.

Also, $\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - ...$

So $x \cos x = x - \frac{x^3}{2} + \frac{x^5}{24} - ...$

Let me use sympy to compute the Taylor expansion and find $a$ and $b$. [/THOUGHT] [PYTHON] from sympy import symbols, integrate, cos, series, limit, ln, simplify, Rational

x, t, a, b = symbols('x t a b')

Compute the integral

integrand = 1 / (1 - t**2) integral = integrate(integrand, (t, 0, x)) print("Integral:", integral)

The expression inside the limit

expr = (a/2) * integral + b * x * cos(x) print("Expression:", expr)

Divide by x^3

expr_over_x3 = expr / x**3

Expand as a series around x = 0

series_expansion = series(expr_over_x3, x, 0, n=4) print("Series expansion:", series_expansion) [/PYTHON]

[THOUGHT] From the series expansion, I can see that: $\frac{1}{x^3}\left(\frac{a}{2}\int_0^x \frac{1}{1-t^2}dt + bx\cos x\right) = \frac{a/2 + b}{x^2} + \frac{a}{6} - \frac{b}{2} + O(x^2)$

For the limit to exist and equal 2 as $x \to 0$, the coefficient of $\frac{1}{x^2}$ must be zero (otherwise the limit would be infinite), and the constant term must equal 2.

So I have two equations:

1. $\frac{a}{2} + b = 0$ (coefficient of $1/x^2$ must be 0)
2. $\frac{a}{6} - \frac{b}{2} = 2$ (constant term equals 2)

From equation 1: $b = -\frac{a}{2}$

Substituting into equation 2: $\frac{a}{6} - \frac{1}{2}\left(-\frac{a}{2}\right) = 2$ $\frac{a}{6} + \frac{a}{4} = 2$ $\frac{2a + 3a}{12} = 2$ $\frac{5a}{12} = 2$ $a = \frac{24}{5}$

Then $b = -\frac{a}{2} = -\frac{12}{5}$

So $a + b = \frac{24}{5} - \frac{12}{5} = \frac{12}{5}$

Therefore $100(a + b)^2 = 100 \cdot \left(\frac{12}{5}\right)^2 = 100 \cdot \frac{144}{25} = \frac{14400}{25} = 576$ [/THOUGHT]

\boxed{576}