<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] A similar approach to @Quanto's is to define $I_n:=\int_0^1\frac{x^ndx}{1+x^2}$ so $I_0=\frac{\pi}{4}$ and $I_n+I_{n+2}=\frac{1}{n+1}$, and in particular $I_n-I_{n+4}=\frac{2}{(n+1)(n+3)}$. Hence $I_4=\frac{\pi}{4}-\frac{2}{3}$ and$$I_8+I_4-4(I_5+I_7)+6I_6=\underbrace{-\frac{2}{35}}_{I_8-I_4}+6\cdot\underbrace{\frac15}_{I_4+I_6}-4I_4-\frac{4}{6}\\=-\frac{2}{35}+\frac65-\pi+\frac83-\frac23=\frac{22}{7}-\pi.$$
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