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### Solution

\[\int_{0}^{1} \frac{x^{4}(1-x)^{4}}{1+x^{2}} d x=?\]

The integral is well defined. Next, let’s use the binomial theorem

\[(1-x)^{4}=x^{4}-4 x^{3}+6 x^{2}-4 x+1\]

\[x^{4}(1-x)^{4}=x^{8}-4 x^{7}+6 x^{6}-4 x^{5}+x^{4}\]

Now, let’s use polynomial division, we get

\[\frac{x^{4}(1-x)^{4}}{1+x^{2}}=x^{6}-4 x^{5}+5 x^{4}-4 x^{2}+4-\frac{4}{x^{2}+1}\]

\[=\frac{1}{7}-\frac{4}{6}+1-\frac{4}{3}+4-4\left(\frac{\pi}{4}\right)-(0-0+0-0+0-4(0))\]

\[=\frac{22}{7}-\pi\]

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