<s>[INST] Alternative way to calculate $\int_0^1(x^4(1-x)^4)/(1+x^2)dx$ $$I=\int_0^1(x^4(1-x)^4)/(1+x^2)dx$$
$$=\int_0^1(x^8-4x^7+6x^6-4x^5+x^4)/(x^2+1)dx$$
$$=\int_0^1(x^6-4x^5+5x^4-4x^2-4/(x^2+1)+4)dx$$
$$=[1/7x^7-2/3x^6+x^5-4/3x^3-4\tan^{-1}x+4x]_0^1$$
$$I=22/7-\pi$$
Any other method to solve this problem?
 [/INST] Here is a systematic procedure. Let $\tan t=x$,
$$I=\int_0^1(x^4(1-x)^4)/(1+x^2)dx=\int_0^{\pi/4}\tan^4t(1-\tan t)^4dt$$
$$=I(8) -4I(7)+6I(6)-4I(5)+I(4)\tag{1}$$
where $I(n)=\int_0^{\pi/4}\tan^nt\>dt$ and it has the recursive relationship $I(n)=\frac{1}{n-1}-I(n-2)$. 
Now, use the recursive equation to get 
$$4I(7)+4I(5)=\frac 23$$
$$I(8)+6I(6)+I(4)=\frac87-4I(4)$$
$$I(4) = -\frac23 +\frac\pi4$$
Then, the integral evaluates to,
$$I=\frac{8}{7}-4(-\frac23 +\frac\pi4) -\frac{2}{3}= \frac{22}{7}-\pi$$
</s>