Definite Integral Evaluation of Rational Function

To evaluate the definite integral
$I = \int_0^1 \frac{x^4(1-x)^4}{1+x^2} \, dx$

First, we expand the numerator $x^4(1-x)^4$:
$x^4(1-x)^4 = x^4(1 - 4x + 6x^2 - 4x^3 + x^4) = x^8 - 4x^7 + 6x^6 - 4x^5 + x^4$

Next, we perform polynomial long division of the numerator $x^8 - 4x^7 + 6x^6 - 4x^5 + x^4$ by $x^2 + 1$:
$\frac{x^8 - 4x^7 + 6x^6 - 4x^5 + x^4}{x^2 + 1} = x^6 - 4x^5 + 5x^4 - 4x^2 + 4 - \frac{4}{1+x^2}$

Now, we can integrate the expression term-by-term:
$I = \int_0^1 \left( x^6 - 4x^5 + 5x^4 - 4x^2 + 4 - \frac{4}{1+x^2} \right) dx$

$I = \left[ \frac{x^7}{7} - \frac{4x^6}{6} + \frac{5x^5}{5} - \frac{4x^3}{3} + 4x - 4\tan^{-1}(x) \right]_0^1$

$I = \left[ \frac{x^7}{7} - \frac{2x^6}{3} + x^5 - \frac{4x^3}{3} + 4x - 4\tan^{-1}(x) \right]_0^1$

Substituting the upper limit $x = 1$ and lower limit $x = 0$:
$I = \left( \frac{1}{7} - \frac{2}{3} + 1 - \frac{4}{3} + 4 - 4\tan^{-1}(1) \right) - 0$

$I = \frac{1}{7} - 2 + 5 - 4\left(\frac{\pi}{4}\right)$

$I = \frac{1}{7} + 3 - \pi$

$I = \frac{22}{7} - \pi$

Hence, the correct option is A.