<s>[INST] Values of this integeration The values of $$I=\int_{0}^{1} {x^4} ({1-x})^4\dfrac1{1+x^2}dx$$
My Failed Attempt 
I tried applying F(x)  =F(a+b-x)  in hopes the denominator will cancel out but it's not. I tried to further simplify the expression but with no luck 
$$I=\int_{0}^{1} {x^4} ({1-x})^4\dfrac1{1+({1-x})^2}dx$$
Adding both integral isn't helping me much 
 [/INST] $$\int_{0}^1 x^4(1-x)^4\dfrac{1}{1+x^2}$$
First compute the integral without boundaries $$\int x^4(1-x)^4\dfrac{1}{1+x^2}$$
Apply long division on $\dfrac{x^4(1-x)^4}{x^2+1}$ and we get in the form of $x^6-4x^5+5x^4-4x^2+4-\dfrac{4}{x^2+1}$
Now you can easily compute the integral $$\int x^6-4x^5+5x^4-4x^2+4-\dfrac{4}{x^2+1}\ dx$$
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