$$I=\int_{0}^{1}\frac{x^4(1-x)^4}{1+x^2}dx$$ I was recently in a small, friendly integration competition and this problem came up. Eventually, I found a solution using messy long division (shown below). I was wondering of other methods (preferably real-method solutions, but complex methods such as residues (if applicable) would also be nice to see) that could be used here.
I spent a lot of time trying to find a nice trigonometric substitution that would work (if you use $x=\tan(u)$, it can be brought to integrating $\int_{0}^{\frac{\pi}{4}}\tan^4(u)(1-\tan(u))^4dx$ which doesn't seem fun/faster and other substitutions I tried ended up similarly). So, that's what I'm hoping to see. There may also be a clever change of variables that that I didn't see (e.g. somehow using $\int_{a}^{b} f(x)dx=\int_{a}^{b}f(a+b-x)dx$ or similar)
Long Division Method:
Multiply the numerator out to prepare for long division: $$(1-x)^4=((1-x)^2)^2=(1-2x+x^2)^2=1-4x+6x^2-4x^3+x^4$$ hence $$x^4(1-x)^4=x^4-4x^5+6x^6-4x^7+x^8$$ Then long division shows that $$\dfrac{x^4(1-x)^4}{1+x^2}=x^6-4x^5+5x^4-4x^2+4-4\cdot\dfrac{1}{1+x^2}$$ Finally, we get that $$I=\int_{0}^{1}\left[x^6-4x^5+5x^4-4x^2+4-4\cdot\frac{1}{1+x^2}\right]dx$$ $$I=\frac{1}{7}x^7-\frac{2}{3}x^6+x^5-\frac{4}{3}x^3+4x-4\arctan(x)\bigg\vert_{0}^{1}$$ $$I=\frac{1}{7}-\frac{2}{3}+1-\frac{4}{3}+4-4\cdot\frac{\pi}{4}-0=\frac{1}{7}+3-\pi=\boxed{\frac{22}{7}-\pi}$$ (Yes, I do see the cleverness of $\frac{22}{7}$, an approximation of $\pi$.)
