<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] Not really that different, but this avoids doing most of the multiplications. It is probably a bit longer though.
First a simplification of your computations. Write $x^4=x^4-1+1$. Then
$$\frac{x^4(1-x)^4}{1+x^2}=\frac{(x^4-1)(1-x)^4}{1+x^2}+\frac{(1-x)^4}{1+x^2}=(x^2+1)(1-x)^4+\frac{(1-x)^4}{1+x^2}$$
Now, do the substitution $u=1-x$. Then
$$I=\int_0^1(x^4(1-x)^4)/(1+x^2)dx= \int_0^1(x^2+1)(1-x)^4 dx +\int_0^1 \frac{(1-x)^4}{1+x^2} d x  \\
= \int_0^1(u^2-2u+1)u^4 d u+ \int_0^1 \frac{(1-2x+x^2)^2}{1+x^2} d x $$
The first integral is trivial, while the second can be simplified the following way:
$$\frac{(1-2x+x^2)^2}{1+x^2} = \frac{1-2x+x^2}{1+x^2} (1-2x+x^2) =\left(1- \frac{2x}{1+x^2}\right)) (1-2x+x^2)\\=(1-2x+x^2)-(2x) \frac{1-2x+x^2}{1+x^2}=(1-2x+x^2)-(2x) \left(1- \frac{2x}{1+x^2}\right)
\\=(1-2x+x^2)-(2x) + \frac{4x^2}{1+x^2}=1-4x+x^2 + 4-\frac{4}{1+x^2}$$
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