Integrate It!- Part V

The given integral is one of the proofs that $\large\frac{22}{7}>\pi$ : $\begin{aligned} S=\int_0^1\frac{x^4(1-x)^4}{1+x^2}\, dx&>0\\ \int_0^1\frac{x^4-4x^5+6x^6-4x^7+x^8}{1+x^2}\, dx&>0\\ \int_0^1\left[x^6-4x^5+5x^4-4x^2+4-\frac{4}{1+x^2}\right]\, dx&>0\\ \left[\frac{x^7}{7}-\frac{2x^6}{3}+x^5-\frac{4x^3}{3}+4x-4\arctan x\right]_0^1&>0\\ \frac{1}{7}-\frac{2}{3}+1-\frac{4}{3}+4-\pi&>0\\ S=\frac{22}{7}-\pi&>0\\ \frac{22}{7}&>\pi \end{aligned}$ So $\left[10000\left(\frac{22}{7}-\pi\right)\right]=12$

Source : The first problem in the 1968 Putnam Competition

$S=\int _{ 0 }^{ 1 }{ \frac { { x }^{ 4 } }{ 1+{ x }^{ 2 } } { (1-x) }^{ 4 } } dx\\ =\int _{ 0 }^{ 1 }{ \frac { { x }^{ 4 } }{ 1+{ x }^{ 2 } } { (1-2x+{ x }^{ 2 }) }^{ 2 } } dx\\ =\int _{ 0 }^{ 1 }{ \frac { { x }^{ 4 } }{ 1+{ x }^{ 2 } } { ({ (1+{ x }^{ 2 }) }^{ 2 }-4x(1+{ x }^{ 2 })+4{ x }^{ 2 }) }^{ 1 } } dx\\ =\int _{ 0 }^{ 1 }{ ({ x }^{ 6 } } -4{ x }^{ 5 }+{ x }^{ 4 }+4\frac { { x }^{ 6 } }{ 1+{ x }^{ 2 } } )dx\\ =\int _{ 0 }^{ 1 }{ ({ x }^{ 6 } } -4{ x }^{ 5 }+{ x }^{ 4 }+4({ x }^{ 4 }-{ x }^{ 2 }+1-\frac { 1 }{ 1+{ x }^{ 2 } } )dx\\ Solving\quad this\quad this\quad we\quad get\quad S=\frac { 22 }{ 7 } -\pi \quad which\quad is\quad not\quad equal\quad to\quad zero\quad as\quad exact\quad value\quad of\quad \pi \neq \frac { 22 }{ 7 }$ $\\ S=0.001264\quad [10000S]=12\\$