The user wants me to solve the definite integral: $I = \int\limits_0^1 {{{{x^4}{{\left( {1 - x} \right)}^4}} \over {1 + {x^2}}}dx}$

This is a well-known integral related to an approximation of $\pi$. Let's solve it from first principles.

Step 1: Expand the numerator. The numerator is $N(x) = x^4(1-x)^4 = (x(1-x))^4 = (x-x^2)^4$. Using the binomial expansion for $(a-b)^4 = a^4 - 4a^3b + 6a^2b^2 - 4ab^3 + b^4$. Here, $a=x$ and $b=x^2$. So, $N(x) = (x)^4 - 4(x)^3(x^2) + 6(x)^2(x^2)^2 - 4(x)(x^2)^3 + (x^2)^4$ $N(x) = x^4 - 4x^5 + 6x^6 - 4x^7 + x^8$.

So the integral becomes: $I = \int_0^1 \frac{x^8 - 4x^7 + 6x^6 - 4x^5 + x^4}{1+x^2} dx$

Step 2: Perform polynomial long division. We need to divide the numerator polynomial $P(x) = x^8 - 4x^7 + 6x^6 - 4x^5 + x^4$ by the denominator polynomial $D(x) = x^2 + 1$.

Performing the long division:

        x^6 - 4x^5 + 5x^4 - 4x^2 + 4      _________________________________x^2+1 | x^8 - 4x^7 + 6x^6 - 4x^5 + x^4       -(x^8         + x^6)      _________________________________            -4x^7 + 5x^6 - 4x^5           -(-4x^7         - 4x^5)           _________________________                   5x^6 + 0x^5 + x^4                  -(5x^6         + 5x^4)                  _________________________                          -4x^4                         -(-4x^4         - 4x^2)                         _________________________                                  4x^2                                 -(4x^2 + 4)                                 ____________                                        -4

So, the division gives a quotient $Q(x) = x^6 - 4x^5 + 5x^4 - 4x^2 + 4$ and a remainder $R(x) = -4$. Therefore, the integrand can be written as: $\frac{x^8 - 4x^7 + 6x^6 - 4x^5 + x^4}{1+x^2} = (x^6 - 4x^5 + 5x^4 - 4x^2 + 4) - \frac{4}{1+x^2}$

Step 3: Integrate the resulting expression. The integral now is: $I = \int_0^1 \left( x^6 - 4x^5 + 5x^4 - 4x^2 + 4 - \frac{4}{1+x^2} \right) dx$

Let's integrate term by term: $I = \left[ \frac{x^7}{7} - 4\frac{x^6}{6} + 5\frac{x^5}{5} - 4\frac{x^3}{3} + 4x - 4 \tan^{-1}(x) \right]_0^1$ $I = \left[ \frac{x^7}{7} - \frac{2x^6}{3} + x^5 - \frac{4x^3}{3} + 4x - 4 \tan^{-1}(x) \right]_0^1$

Step 4: Evaluate the definite integral. Now, substitute the limits of integration. The value at the lower limit $x=0$ is 0 for all terms. So, we just need to evaluate at the upper limit $x=1$.

$I = \left( \frac{1^7}{7} - \frac{2(1)^6}{3} + 1^5 - \frac{4(1)^3}{3} + 4(1) - 4 \tan^{-1}(1) \right) - (0)$ $I = \frac{1}{7} - \frac{2}{3} + 1 - \frac{4}{3} + 4 - 4 \left(\frac{\pi}{4}\right)$ $I = \frac{1}{7} - \left(\frac{2}{3} + \frac{4}{3}\right) + (1+4) - \pi$ $I = \frac{1}{7} - \frac{6}{3} + 5 - \pi$ $I = \frac{1}{7} - 2 + 5 - \pi$ $I = \frac{1}{7} + 3 - \pi$

Now, combine the fractions: $I = \frac{1 + 3 \times 7}{7} - \pi$ $I = \frac{1 + 21}{7} - \pi$ $I = \frac{22}{7} - \pi$

Step 5: Compare with the options. The calculated value is $I = \frac{22}{7} - \pi$. Let's check the given options: A: $\frac{22}{7} - \pi$ B: $\frac{2}{105}$ C: $0$ D: $\frac{71}{15} - \frac{3\pi}{2}$

Our result matches option A exactly.