<s>[INST] 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
 [/INST] Note that $(1 \pm i)^4  = -4$, hence $(1-x)^4 + 4 = p(x)(1+x^2)$ for some normalized second degree polynomial $p$. To find $p$, we plug in $0$, $1$, giving 
$$ 5 = p(0), 4 = 2p(1) $$
With the ansatz $p(x) = x^2 + ax + b$, we have
$$ 5 = b,\quad 2+2a + 2b = 4 \iff a = -4 $$
Hence
$$ \frac{(1-x)^4}{1+x^2} = x^2 - 4x + 5 - \frac{4}{1+x^2} $$
Multiplying by $x^4$ gives 
$$ \frac{x^4(1-x)^4}{1+x^2} = (x^6 - 4x^5 + 5x^4) - \frac{4x^4}{1+x^2} $$
Now we use
$$ \frac{x^4}{1+x^2} = \frac{x^4 - 1}{x^2 + 1} + \frac{1}{x^2+ 1} = x^2 - 1 + \frac{1}{x^2 + 1}$$
giving
$$ \frac{x^4(1-x)^4}{1+x^2} = x^6 - 4x^5 + 5x^4 - 4x^2 + 4 - \frac 4{x^2 + 1} $$
So
$$ \int \frac{x^4(1-x)^4}{1+x^2}\, dx = \frac{x^7}7  - \frac 23x^6 + x^5 - \frac 43x^3 + 4x - 4\arctan x. $$
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