<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] Literally all you have to do is expand it out, divide, and integrate.
$$
x^4(1-x)^4 = x^8 - 4x^7 + 6x^6 - 4x^5 + x^4
$$
Now use long division:
$$
x^4(1-x)^4 = x^6(x^2+1)- 4x^5(x^2+1) + 5x^4(x^2 + 1) - 4x^2(x^2+1)+4(x^2+1) -4 \\
\frac{x^4(1-x)^4}{1+x^2} = (x^6 -4x^5 + 5x^4 - 4x^2 + 4) - \frac{4}{x^2+1} \\
$$
We integrate this:
$$
\int_0^1 \! (x^6 -4x^5 + 5x^4 - 4x^2 + 4) \,\mathrm{d}x - \int_0^1 \! \frac{4}{x^2+1} \, \mathrm{d}x \\
= \frac{22}{7} - 4(\arctan(1) - \arctan(0)) \\
= \frac{22}{7} - \pi
$$
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