$\displaystyle\int_0^1 \dfrac{x^4(1-x)^4}{1 + x^2}\, dx = ?$ (Putnam — pinpoints $22/7 - \pi$)

Polynomial-divide $x^4(1-x)^4 = x^8 - 4x^7 + 6x^6 - 4x^5 + x^4$. Divide by $1 + x^2$: quotient $x^6 - 4x^5 + 5x^4 - 4x^2 + 4$ with remainder $-4$. So integrand = quotient $- 4/(1+x^2)$. Integrate $0$ to $1$: $1/7 - 4/6 + 1 - 4/3 + 4 - 4\cdot\pi/4 = 1/7 - 2/3 + 1 - 4/3 + 4 - \pi = 22/7 - \pi$.