$$\int\limits_0^1 {{{{x^4}{{(1 - x)}^4}} \over {1 + {x^2}}}dx = \int\limits_0^1 {{{{x^4}{{\{ (1 + {x^2}) - 2x\} }^2}} \over {1 + {x^2}}}dx} } $$

$$ = \int\limits_0^1 {{x^4}{{{{(1 + {x^2})}^2} - 4x(1 + {x^2}) + 4{x^2}} \over {1 + {x^2}}}dx} $$

$$ = \int\limits_0^1 {{x^4}\left[ {1 + {x^2} - 4x + {{4\{ 1 + {x^2} - 1\} } \over {1 + {x^2}}}} \right]dx} $$

$$ = \int\limits_0^1 {{x^4}\left[ {1 + {x^2} - 4x + 4 - {4 \over {1 + {x^2}}}} \right]dx} $$

$$ = \int\limits_0^1 {\left[ {{x^6} - 4{x^5} + 5{x^4} - 4{{{x^4} - 1 + 1} \over {1 + {x^2}}}} \right]dx} $$

$$ = \int\limits_0^1 {({x^6} - 4{x^5} + 5{x^4} - 4{x^2} + 4)dx - 4\int\limits_0^1 {{{dx} \over {1 + {x^2}}}} } $$

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

$$ = \left[ {{1 \over 7} - {2 \over 3} + 1 - {4 \over 3} + 4} \right] - \pi = {{22} \over 7} - \pi $$