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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?

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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 $10/3-\pi$.

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  • $\begingroup$ Thanks! I wanted just that: a new creative approach to learn some good technique. The speed isn't a problem,if it was, I would have done it with wolfram. (Before accept your answer I'll wait to see if there's some other ) $\endgroup$ – Collapsed Jan 16 '16 at 20:00
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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.

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