Shortest method for $\int_{0}^{1}\frac{x^{4}\left(1-x\right)^{4}}{1+x^{2}}$ I don't want to solve by expanding it and all, I tried corollary but denominator becomes messy, also in the options there is $\pi$ so tried several trigonometric substitutions too

$f(z)=\frac{z^4(1-z)^4}{1+z^2}$ is a meromorphic function with simple poles at $z=\pm i$. Since
$$\text{Res}_{z=\pm i}\frac{z^4(1-z)^4}{(z+i)(z-i)}=\lim_{z\to \pm i}\frac{z^4(1-z)^4}{z\pm i}=\pm 2i \tag{1}$$
we have that
$$ f(z)-\left(\frac{2i}{z-i}+\frac{-2i}{z+i}\right) = f(z)+\frac{4}{z^2+1}\tag{2} $$
is an entire function: a polynomial, in particular. It follows that $z^4(1-z)^4+4$ is a multiple of $z^2+1$. Not surprising, since by Sophie Germain's identity
$$ w^4+4 = (w^2-2w+2)(w^2+2w+2) \tag{3}$$
hence:
$$ z^4(1-z)^4+4 = (1+z^2)(2-2z+z^2)(2+2 z-z^2-2 z^3+z^4)\tag{4} $$
as well as:
$$ \int_{0}^{1}\frac{x^4(1-x)^4}{1+x^2}\,dx = -\pi+\int_{0}^{1}(2-2z+z^2)(2+2 z-z^2-2 z^3+z^4)\,dz.\tag{5}$$
