Let $ a$ be positive constant number. Evaluate $ \int_{ \minus{} a}^a \frac {x^2\cos x \plus{} e^{x}}{e^{x} \plus{} 1}\ dx.$
1. Consider the integral \( I = \int_{-a}^a \frac{x^2 \cos x + e^x}{e^x + 1} \, dx \).

2. We can split the integral into two parts:
   \[
   I = \int_{-a}^0 \frac{x^2 \cos x + e^x}{e^x + 1} \, dx + \int_0^a \frac{x^2 \cos x + e^x}{e^x + 1} \, dx
   \]

3. To simplify the integral, we use the property of definite integrals and the symmetry of the integrand. Let \( f(x) = \frac{x^2 \cos x + e^x}{e^x + 1} \). Notice that:
   \[
   f(-x) = \frac{(-x)^2 \cos(-x) + e^{-x}}{e^{-x} + 1} = \frac{x^2 \cos x + e^{-x}}{e^{-x} + 1}
   \]

4. We can rewrite the integral using the property of definite integrals:
   \[
   I = \int_{-a}^a f(x) \, dx = \int_{-a}^a \frac{x^2 \cos x + e^x}{e^x + 1} \, dx
   \]

5. By the property of definite integrals, we have:
   \[
   \int_{-a}^a f(x) \, dx = \int_{-a}^a \frac{x^2 \cos x + e^x}{e^x + 1} \, dx = \int_{-a}^a \frac{x^2 \cos x + e^{-x}}{e^{-x} + 1} \, dx
   \]

6. Adding these two integrals, we get:
   \[
   2I = \int_{-a}^a \left( \frac{x^2 \cos x + e^x}{e^x + 1} + \frac{x^2 \cos x + e^{-x}}{e^{-x} + 1} \right) \, dx
   \]

7. Simplify the integrand:
   \[
   \frac{x^2 \cos x + e^x}{e^x + 1} + \frac{x^2 \cos x + e^{-x}}{e^{-x} + 1} = x^2 \cos x + 1
   \]

8. Therefore, we have:
   \[
   2I = \int_{-a}^a (x^2 \cos x + 1) \, dx
   \]

9. Split the integral:
   \[
   2I = \int_{-a}^a x^2 \cos x \, dx + \int_{-a}^a 1 \, dx
   \]

10. The integral of \(1\) over \([-a, a]\) is:
    \[
    \int_{-a}^a 1 \, dx = 2a
    \]

11. The integral of \(x^2 \cos x\) over \([-a, a]\) is zero because \(x^2 \cos x\) is an odd function and the integral of an odd function over a symmetric interval is zero:
    \[
    \int_{-a}^a x^2 \cos x \, dx = 0
    \]

12. Therefore:
    \[
    2I = 2a
    \]

13. Solving for \(I\):
    \[
    I = a
    \]

The final answer is \(\boxed{a}\)