Help on a very interesting integral: $\int_0^1\frac{x^5(1-x)^4}{1+x^3}\,dx$ I've been starting to work on problems with difficult integrals, and I came to
one of my problems where I end up having to integrate
$$\int_0^1 \frac{x^5(1-x)^4}{1+x^3} dx.$$
Wolfram seems to time out on integrals like these, so I decided to tackle this
integral by myself.
It seems like to me that there should be a meaningful integration by parts
that occurs here, in particular a relationship that could be developed between
$x^5$ and $\frac{(1-x)^4}{1+x^3}.$ The problem is, I'm having some difficulties
with the difficulty of building a meaningful integral of $dv = \frac{(1-x)^4}{
1+x^3} dx.$ any recommendations on how to go about starting with direction on
this problem?

In general, any integral of the form $\int_{0}^{1}\frac{p(x)}{1+x^3}\,dx $ with $p(x)\in\mathbb{Q}[x]$ can be written as a linear combination with rational coefficients of $1$ and 
$$ I_1=\int_{0}^{1}\frac{x}{1+x^3}\,dx=\frac{1}{9} \left(\pi \sqrt{3} -3\log 2\right),\qquad I_2= \int_{0}^{1}\frac{x^2}{1+x^3}\,dx=\frac{1}{3}\log(2) $$
$$ I_0=\int_{0}^{1}\frac{1}{1+x^3}\,dx = \frac{1}{9} \left(\pi\sqrt{3} +3\log 2\right)$$
and we just need to perform a polynomial division betwen $p(x)$ and $(1+x^3)$ to compute the original integral. Notice that $I_2$ is trivial and
$$ I_1+I_0 = \int_{0}^{1}\frac{dx}{1-x+x^2}=\frac{2\pi}{3\sqrt{3}}, $$
$$ I_1-I_0 = \int_{0}^{1}\frac{x-1}{x^3+1}=-\sum_{n\geq 0}\left(\frac{1}{6n+1}-\frac{1}{6n+2}-\frac{1}{6n+4}+\frac{1}{6n+5}\right)=-\frac{2\log 2}{3}$$
are quite simple to prove.
