<s>[INST] William Lowell Putnam Integral Problem Prove That
$$
\frac{22}{7}-\pi
=
\int_{0}^{1}\frac{x^{4}\left(1 - x\right)^{4}}{1 + x^{2}}\,{\rm d}x
$$
 [/INST] $$\begin{align}\int_0^1 dx \frac{x^4 (1-x)^4}{1+x^2} &= \int_0^1 dx \frac{x^4-4 x^5 + 6 x^6 - 4 x^7 + x^8}{1+x^2}\\ &= \sum_{k=0}^{\infty} (-1)^k \int_0^1 dx \left (x^{4+2 k}-4 x^{5+2 k}+6 x^{6+2 k}-4 x^{7+2 k}+x^{8+2 k}\right )\\&= \sum_{k=0}^{\infty} (-1)^k \left (\frac1{5+2 k}-\frac{4}{6+2 k}+\frac{6}{7+2 k}-\frac{4}{8+2 k}+\frac1{9+2 k} \right ) \end{align} $$
Now,
$$\sum_{k=0}^{\infty} \frac{(-1)^k}{5+2 k} = \frac15-\frac17+\frac19-\cdots=\frac{\pi}{4}-\left (1-\frac13 \right )  $$
Similarly,
$$\sum_{k=0}^{\infty} \frac{(-1)^k}{7+2 k} = \frac17-\frac19+\frac1{11}-\cdots=\left (1-\frac13+\frac15 \right ) -\frac{\pi}{4} $$
$$\sum_{k=0}^{\infty} \frac{(-1)^k}{9+2 k} = \frac19-\frac1{11}+\frac1{13}-\cdots=\frac{\pi}{4}-\left (1-\frac13+\frac15-\frac17 \right )  $$
Meanwhile,
$$\sum_{k=0}^{\infty} \frac{(-1)^k}{3+ k} = \frac13-\frac14+\frac15-\cdots=\log{2}-\left (1-\frac12 \right )$$
$$\sum_{k=0}^{\infty} \frac{(-1)^k}{4+ k} = \frac14-\frac15+\frac16-\cdots=\left (1-\frac12+\frac13 \right )-\log{2}$$
The result is, summing all of the above numbers,
$$-\left (1-\frac13 \right )+ 6 \left (1-\frac13+\frac15 \right )-\left (1-\frac13+\frac15-\frac17 \right )- \frac{2}{3} - \pi = \frac{22}{7}-\pi$$
</s>