The value(s) of \(\int_0^1 \frac{x^4(1-x)^4}{1+x^2} d x\) is (are)

Let \(I=\int_0^1 \frac{x^4(1-x)^4}{1+x^2} d x\)
\[
\begin{aligned}
& =\int_0^1 \frac{\left(x^4-1\right)(1-x)^4+(1-x)^4}{\left(1+x^2\right)} d x \\
& =\int_0^1\left(x^2-1\right)(1-x)^4 d x \\
& +\int_0^1 \frac{\left(1+x^2-2 x\right)^2}{\left(1+x^2\right)} d x \\
& =\int_0^1\left\{\left(x^2-1\right)(1-x)^4+\left(1+x^2\right)-4 x\right. \\
& \left.+\frac{4 x^2}{\left(1+x^2\right)}\right\} d x \\
& =\int_0^1\left(\left(x^2-1\right)(1-x)^4+\left(1+x^2\right)-4 x\right. \\
& \left.+4-\frac{4}{1-x^2}\right) d x \\
& =\int_0^1\left(x^6-4 x^5+5 x^4-4 x^2\right. \\
& \left.+4-\frac{4}{1+x^2}\right) d x \\
& =\left[\frac{x^7}{7}-\frac{4 x^6}{6}+\frac{5 x^5}{5}\right. \\
&
\end{aligned}
\]

\[
\begin{aligned}
& \left.\quad-\frac{4 x^3}{3}+4 x-4 \tan ^{-1} x\right]_0^1 \\
= & \frac{1}{7}-\frac{4}{6}+\frac{5}{5}-\frac{4}{3}+4-4\left(\frac{\pi}{4}-0\right) \\
= & \frac{22}{7}-\pi
\end{aligned}
\]