Statement - I: The value of the integral ∫π/3π/6 dx/(1+arctanx) is equal to π/6. Statement - II: ∫aaf(x)dx = ∫aaf(a+b-x)dx. Select the correct option regarding the truth values of Statement I and Statement II.

Let I = ∫π/3π/6 dx/(1+arctanx). Let x = π/2 - u. Then dx = -du. When x = π/3, u = π/6. When x = π/6, u = π/3. Therefore, I = ∫π/3π/6 -du/(1+arctan(π/2-u)). Let f(x) = 1/(1+arctan(x)). Then ∫ba f(x) dx = ∫ba f(a+b-x) dx is a property of definite integrals. Let's check statement I. Let I = ∫π/3π/6 dx/(1+arctanx). Let f(x) = 1/(1+arctanx). Then f(π/3) ≈ 0.46 and f(π/6) ≈ 0.58. The average value of f(x) in the interval [π/6, π/3] is approximately 0.52. The width of the interval is π/6. Therefore, the approximate value of the integral is 0.52 * π/6 ≈ 0.27. This is not equal to π/6. Hence, Statement I is False. Statement II is True. Therefore, the correct option is 'Statement -I is False; Statement -II is True'.