Creative way for $\int _0 ^1 \frac{x^4 (1-x)^4} {x^2 +1} {\rm d}x$ I'm able to do this $\int _{0} ^1 \frac{x^4 (1-x)^4} {x^2 +1} {\rm d}x = \frac{22} 7 - \pi$ only with polynomial long division. It's lame.
I tried a more creative way to do it with less algebra, like a substitution with $t= \arctan x - x$ because the derivative is worth it ${\rm d} t = -\frac {x^2} {x^2 +1} {\rm d} x$ but I'm unable to separate the $x$ in the $t = \arctan x - x$. Is there a way to separate the $x$, or a complete different way to do it?
 A: I can suggest the following more creative approach, which avoids polynomial divisions (but is not necessarily faster). I think it might be instructive (at least I had a lot of fun working it out). Write 
$$
\frac{1}{x^2+1}=\int_0^\infty ds\ e^{-s (x^2+1)}\ .
$$
Therefore
$$
\int_0^1dx\frac{x^4(1-x)^4}{x^2+1}=\int_0^\infty ds e^{-s}\frac{\partial^2}{\partial s^2}\int_0^1 dx\ (1-x)^4 e^{-s x^2}\ .
$$
Integrating by parts (twice)
$$
e^{-s}\frac{\partial}{\partial s}\int_0^1 dx\ (1-x)^4 e^{-s x^2}\Big|_{0}^\infty+\int_0^\infty ds e^{-s}\frac{\partial}{\partial s}\int_0^1 dx\ (1-x)^4 e^{-s x^2}
$$
$$
=\int_0^1 dx\ x^2 (1-x)^4-\int_0^1 dx (1-x)^4+\int_0^\infty ds e^{-s}\int_0^1 dx\ (1-x)^4 e^{-s x^2}
$$
The first two integrals are elementary. The third one can be handled and gives $10/3-\pi$.
A: Here is a solution using a substitution, and instead of polynomial long division we end up using a reduction when integrating tangent functions. I don't claim it is easier. It is just (slightly) different.
Substitution
By the substitution $u=\arctan x$ the integral transforms into
$$
\int_0^{\pi/4}\tan^4u(1-\tan u)^4\,du.\tag{*}
$$
Reduction
Next, since $D\tan u=1+\tan^2u$ we find that for $n\geq 2$ (here, we should
interpret $\tan^0u$ as being equal to $1$)
$$
\begin{aligned}
\int \tan^{n}u\,du&=\int\tan^{n-2}u(1+\tan^2u)\,du-\int\tan^{n-2}u\,du\\
&=\frac{1}{n-1}\tan^{n-1}u-\int\tan^{n-2}u\,du.
\end{aligned}
$$
Applying this reduction several times on
$$
\tan^4u(1-\tan u)^4=\tan^8u-4\tan^7u+6\tan^6u-4\tan^5u+\tan^4u
$$
we end up finding that the integral $(*)$ equals

$$\Bigl[\frac{1}{7}\tan^7u-\frac{2}{3}\tan^6u+\tan^5u-\frac{4}{3}\tan^3u+4\tan u-4u\Bigr]_0^{\pi/4}=\frac{22}{7}-\pi.$$

Here, it could be funny to note that while reducing the odd powers of $\tan u$, we do not get any $\int\tan u\,du$, a term which would have given us logarithms in the answer.
