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?
 A: Here is a systematic procedure. Let $\tan t=x$,
$$I=\int_0^1(x^4(1-x)^4)/(1+x^2)dx=\int_0^{\pi/4}\tan^4t(1-\tan t)^4dt$$
$$=I(8) -4I(7)+6I(6)-4I(5)+I(4)\tag{1}$$
where $I(n)=\int_0^{\pi/4}\tan^nt\>dt$ and it has the recursive relationship $I(n)=\frac{1}{n-1}-I(n-2)$. 
Now, use the recursive equation to get 
$$4I(7)+4I(5)=\frac 23$$
$$I(8)+6I(6)+I(4)=\frac87-4I(4)$$
$$I(4) = -\frac23 +\frac\pi4$$
Then, the integral evaluates to,
$$I=\frac{8}{7}-4(-\frac23 +\frac\pi4) -\frac{2}{3}= \frac{22}{7}-\pi$$
A: 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:
$$\frac{(1-2x+x^2)^2}{1+x^2} = \frac{1-2x+x^2}{1+x^2} (1-2x+x^2) =\left(1- \frac{2x}{1+x^2}\right)) (1-2x+x^2)\\=(1-2x+x^2)-(2x) \frac{1-2x+x^2}{1+x^2}=(1-2x+x^2)-(2x) \left(1- \frac{2x}{1+x^2}\right)
\\=(1-2x+x^2)-(2x) + \frac{4x^2}{1+x^2}=1-4x+x^2 + 4-\frac{4}{1+x^2}$$
A: 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$$I_8+I_4-4(I_5+I_7)+6I_6=\underbrace{-\frac{2}{35}}_{I_8-I_4}+6\cdot\underbrace{\frac15}_{I_4+I_6}-4I_4-\frac{4}{6}\\=-\frac{2}{35}+\frac65-\pi+\frac83-\frac23=\frac{22}{7}-\pi.$$
