<s>[INST] Finding $\int_0^1{\frac{x^4(1-x)^4}{1+x^2}}dx$ The question I am working on: Evaluate 
$$\frac{1}{2} \int^1_0{x^4 (1-x)^4 } dx \le \int^1_0{\frac{x^4 (1-x)^4}{1+x^2}} dx \le \int^1_0{x^4 (1-x)^4 } dx$$
So using integration by parts to solve: 
(letting $u=(1-x)^4$ and $dv=x^4$)
$$\int{x^4 (1-x)^ 4} dx = \frac{4}{5}x^5(x-1) - \frac{4}{5}\left(\frac{x^7}{7} - \frac{x^6}{6}\right)+c$$
Is it correct so far? If so ... 
$$\int^1_0{ x^4 (1-x)^4 } dx = \frac{4}{5}\left(\frac{1}{6}-\frac{1}{7}\right)=\frac2{105}$$
$$\frac{1}{2} \int^1_0{ x^4 (1-x)^4 } dx = \frac{2}{5}\left(\frac{1}{6}-\frac{1}{7}\right)=\frac1{105}$$
But 
$$\int\frac{{x^4(1-x)^4}}{1+x^2} dx = ??$$

Since I found $\int{{x^4(1-x)^4}} dx$. I thought of integration by parts, letting $u=\frac{1}{1+x^2}$, $dv = x^4(1-x)^4$. But I will get a very complicated $v$ to integrate later? Same if I did it the other way around? 
 [/INST] Hint :
Rewrite integral into form :
$$I=\int \frac{x^8+x^6}{1+x^2} \,dx + \int \frac{x^6+x^4}{1+x^2} \,dx +4\cdot \int \frac{x^6}{1+x^2} \,dx -4\cdot \int \frac{x^7+x^5}{1+x^2} \,dx$$
for third integral do long division .
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