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?

Using the binomial theorem, expand $(1-x)^4$ in the numerator:
$$
I = \int_0^1 \frac{x^5(1-x)^4}{1+x^3}\,dx = \int_0^1 \frac{x^5(1 - 4x + 6x^2 - 4x^3 + x^4)}{1+x^3}\,dx \\
= \int_0^1 \frac{x^5 - 4x^6 + 6x^7 - 4x^8 + x^9}{1+x^3}\,dx.
$$
Using polynomial long division,
$$
I = \int_0^1 \left(x^6 - 4x^5 + 6x^4 - 5x^3 + 5x^2 - 6x + 5 + \frac{-5x^2 + 6x - 5}{x^3 + 1}\right)\,dx.
$$
At this point let's focus on the term $\frac{-5x^2 + 6x - 5}{x^3 + 1}$. Notice that $x^3 + 1 = (x+1)(x^2-x+1),$ and perform partial fraction decomposition:
$$
\int_0^1\frac{-5x^2 + 6x - 5}{x^3 + 1}\,dx = \int_0^1\left(\frac{x+1}{3(x^2-x+1)} - \frac{16}{3(x+1)}\right)\,dx.
$$
Put $\alpha = 3/2$, $\beta^2 = \alpha^2/3$, $u = x-1/2$, and let's focus on the term $\frac{x+1}{3(x^2-x+1)}$. Completing the square in the denominator:
$$
\int_0^1\frac{(x-\tfrac{1}{2}) + \tfrac{3}{2}}{3(x-\tfrac{1}{2})^ 2 + \tfrac{9}{4}}\,dx = \int_{-1/2}^{1/2}\frac{u + \alpha}{3u^2 + \alpha^2}\,du = \frac{1}{3}\int_{-1/2}^{1/2}\frac{\alpha}{u^2 + \tfrac{\alpha^2}{3}}\,du \\= \frac{2\alpha}{3}\int_{0}^{1/2}\frac{1}{u^2 + \beta^2}\,du =  \frac{1}{\beta}\,\arctan{\frac{u}{\beta}}\,\Big|_{0}^{1/2} = \frac{2}{\sqrt{3}}\,\arctan{\frac{1}{\sqrt{3}}} = \frac{2}{\sqrt{3}}\frac{\pi}{6} = \frac{\pi}{3\sqrt{3}},
$$
where we made use of the facts that $2\alpha/3 = 1$, $f(u) = u/(3u^2 + \alpha^2)$ is odd, $g(u) = 1/(3u^2+\alpha^2)$ is even, and our region of integration is symmetric about $u = 0$.
The integral $\int_0^1 \frac{16}{3(x+1)}\,dx$ should be straightforward, and of course the integral of the polynomial portion is easy too.
