[THOUGHT] We need to compute the limit: [ \lim_{x \to 0} \frac{1}{x^3} \left( \frac{a}{2} \int_0^x \frac{1}{1-t^2}, dt + bx\cos x \right) = 2. ] Let's denote: [ F(x) = \frac{a}{2} \int_0^x \frac{1}{1-t^2}, dt + bx\cos x. ] We need to expand F(x) as a power series around x=0 up to order x^3, because the limit is of the form F(x)/x^3 → 2 as x→0. That means F(x) must behave like 2x^3 + o(x^3) near 0. So we need to find a and b such that the coefficients of x^0, x^1, x^2 vanish, and the coefficient of x^3 equals 2.

First, compute the integral: [ \int_0^x \frac{1}{1-t^2}, dt = \frac12 \ln\left(\frac{1+x}{1-x}\right). ] But we can also expand it as a series. Alternatively, we can expand the integrand as a geometric series: [ \frac{1}{1-t^2} = 1 + t^2 + t^4 + \dots \quad \text{for } |t|<1. ] Integrate termwise: [ \int_0^x (1 + t^2 + t^4 + \dots) dt = x + \frac{x^3}{3} + \frac{x^5}{5} + \dots = \sum_{n=0}^\infty \frac{x^{2n+1}}{2n+1}. ] Thus, [ \int_0^x \frac{1}{1-t^2}, dt = x + \frac{x^3}{3} + \frac{x^5}{5} + O(x^7). ] Now, [ \frac{a}{2} \int_0^x \frac{1}{1-t^2}, dt = \frac{a}{2} \left( x + \frac{x^3}{3} + \frac{x^5}{5} + \dots \right) = \frac{a}{2}x + \frac{a}{6}x^3 + O(x^5). ] Next, expand bx cos x: [ \cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} + O(x^6). ] So, [ bx\cos x = bx \left( 1 - \frac{x^2}{2} + \frac{x^4}{24} + \dots \right) = bx - \frac{b}{2}x^3 + \frac{b}{24}x^5 + O(x^7). ] Now sum: [ F(x) = \left( \frac{a}{2}x + \frac{a}{6}x^3 \right) + \left( bx - \frac{b}{2}x^3 \right) + O(x^5). ] Combine like terms: Coefficient of x: (\frac{a}{2} + b). Coefficient of x^3: (\frac{a}{6} - \frac{b}{2}).

For the limit to exist (finite), the coefficient of x must be zero, otherwise F(x) ~ C x, then F(x)/x^3 ~ C/x^2 → ∞ (unless C=0). So we need: [ \frac{a}{2} + b = 0. \quad (1) ] Then the coefficient of x^3 becomes the leading term. The limit is then: [ \lim_{x \to 0} \frac{F(x)}{x^3} = \frac{a}{6} - \frac{b}{2}. ] We are given that this limit equals 2: [ \frac{a}{6} - \frac{b}{2} = 2. \quad (2) ] Solve (1) and (2). From (1): b = -a/2. Substitute into (2): [ \frac{a}{6} - \frac{(-a/2)}{2} = \frac{a}{6} + \frac{a}{4} = \frac{2a}{12} + \frac{3a}{12} = \frac{5a}{12} = 2. ] Thus a = 24/5 = 4.8. Then b = -a/2 = -12/5 = -2.4. Now compute (a+b)^2: a+b = 24/5 - 12/5 = 12/5 = 2.4. So (a+b)^2 = (12/5)^2 = 144/25. Then 100(a+b)^2 = 100 * 144/25 = (100/25)144 = 4144 = 576. Thus the answer is 576.

But we should verify with Python to ensure no algebraic mistakes. We'll compute the limit directly using sympy to confirm. [/THOUGHT] [PYTHON] import sympy as sp

Define symbols

x, a, b = sp.symbols('x a b')

Define the integral

integral = sp.integrate(1/(1 - t**2), (t, 0, x))

Define F(x)

F = (a/2) * integral + b * x * sp.cos(x)

Compute limit of F(x)/x^3 as x->0

limit_expr = sp.limit(F / x**3, x, 0) print("Limit expression (symbolic):", limit_expr)

The limit should be an expression in a and b.

Set up equations: coefficient of x^0? Actually limit_expr is a rational expression.

Let's simplify.

limit_simplified = sp.simplify(limit_expr) print("Simplified limit:", limit_simplified)

We know limit = 2.

Solve for a and b.

But we also need condition that lower order terms vanish: i.e., the limit exists.

Actually, sympy's limit will give the value if the limit exists.

Let's compute the series expansion of F(x) up to x^3 to see conditions.

series_F = sp.series(F, x, 0, 4).removeO() print("Series expansion of F(x) up to x^3:", series_F)

The series expansion gives terms: coefficient of x^1 and x^3.

For limit to exist, coefficient of x^1 must be zero.

coeff_x = sp.series(F, x, 0, 2).removeO().coeff(x, 1) print("Coefficient of x:", coeff_x)

Solve coeff_x = 0 and limit_simplified = 2.

sol = sp.solve([coeff_x, limit_simplified - 2], [a, b]) print("Solutions for a, b:", sol)

Compute (a+b)^2 and then 100*(a+b)^2

if sol: a_val = sol[a] b_val = sol[b] expr = 100 * (a_val + b_val)*2 print("100(a+b)^2 =", expr) # Evaluate to integer print("Integer value:", expr.evalf()) [/PYTHON]

[THOUGHT] The Python output shows an error: "NotImplementedError: solving Abs(a + 2*b) when the argument is not real or imaginary." This suggests that the symbolic solving encountered an absolute value due to the limit expression. However, my earlier manual derivation gave a clear result. Since the Python code had issues, I should trust my manual calculation which was verified step-by-step and gave 576. The error is likely due to sympy's handling of the limit expression, but the manual derivation is straightforward and correct. Therefore, the final answer is 576.

[/THOUGHT] \boxed{576}