<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] $\bf{My\; Solution::}$ Let $$\displaystyle I = \int_{0}^{1}\frac{x^4(1-x)^4}{1+x^2}dx\;,$$ Let $\displaystyle x= \tan \phi\;,$ Then $dx = \sec^2 \phi.$
And Changing Limit, We Get $$\displaystyle I = \int_{0}^{\frac{\pi}{4}}\tan^4 \phi\cdot \left(1-\tan \phi\right)^4d\phi.$$
$$\displaystyle  =\int_{0}^{\frac{\pi}{4}}\tan^4 \phi \cdot \left(1-4\tan \phi+6\tan^2 \phi-4\tan^3 \phi+\tan^4 \phi\right).$$
$$\displaystyle = \int_{0}^{\frac{\pi}{4}}\left\{\tan^4 \phi-4\tan^5 \phi+6\tan^6 \phi-4\tan^7 \phi+\tan^8 \phi \right\}d\phi.$$
$$\displaystyle = \int_{0}^{\frac{\pi}{4}}\left\{\left[\tan^4 \phi+\tan^6 \phi \right]-4\left[\tan^5 \phi \cdot \sec^2 \phi \right]+\left[\tan^6 \phi+\tan^8 \phi  \right]+4\tan^6 \phi \right\}d\phi.................(\star)\color{\red}\checkmark.$$
Now Let $$\displaystyle J_{n} = \int_{0}^{\frac{\pi}{4}}\tan^{n}\phi d\phi\;,$$ Then $$\displaystyle J_{n+2} = \int_{0}^{\frac{\pi}{4}}\tan^{n+2}\phi d\phi$$
Now $$\displaystyle J_{n+2}+J_{n} = \int_{0}^{\frac{\pi}{4}}\tan^{n} \phi \cdot \sec^2 \phi d\phi  = \frac{1}{n+1}\Rightarrow J_{n+2}+J_{n} = \frac{1}{n+1}............(\star \star)\color{\red}\checkmark.$$
Now at $n=0\;,$ we get $\displaystyle J_{0} = \frac{\pi}{4}\;,$ Then $\displaystyle J_{2} = \left(1-\frac{\pi}{4}\right)$ and $\displaystyle $ When $ n= 2\;,$ We get $$\displaystyle J_{4} = \left(\frac{\pi}{4}-\frac{2}{3}\right).$$
and When $n=4\;,$ We get $\displaystyle J_{6} = \left(\frac{13}{15}-\frac{\pi}{4}\right).$ and We get $\displaystyle J_{4}+J_{6}=\frac{1}{5}$ And $\displaystyle J_{6}+J_{8} = \frac{1}{7}$ 
Now Put these values in $............(\star)\color{\red}\checkmark.\;,$ We get
$$\displaystyle I = \frac{1}{5}-\frac{2}{3}+\frac{1}{7}+\frac{52}{15}-\pi = \left(\frac{22}{7}-\pi\right).$$
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