You have found, correctly, that
$$\begin{eqnarray*}
I &\equiv& \int_0^\infty\frac{x^{1/3}dx}{1+x^2} \\
&=& \frac{2}{1-e^{2\pi i/3}} \int_{-\infty}^\infty d z\, \frac{z^{5/3}}{1+z^4}.
\end{eqnarray*}$$
The second integral has a branch cut at $z=0$ and singularities at the roots of $1+z^4$.
Close the contour in the upper half-plane.
Let $\gamma$ denote the contour.
We will pick up residues at $e^{i\pi/4}$ and $e^{i 3\pi/4}$.
We deform the contour slightly near $z=0$ to avoid the cut.
Letting $z=\epsilon e^{i\theta}$ we see that the contribution to the integral here goes like
$\epsilon^{1+5/3} = \epsilon^{8/3}$ and so vanishes in the limit.
Letting $z = R e^{i\theta}$, we find the integral over the semicircle at infinity goes like
$R^{1+5/3-4} = 1/R^{4/3}$.
Thus, the contribution from this part of the contour is also zero.
Therefore, $\int_{-\infty}^\infty d z\, z^{5/3}/(1+z^4) = \int_\gamma d z\, z^{5/3}/(1+z^4)$.
We find
$$\begin{eqnarray*}
I &=& \frac{2}{1-e^{2\pi i/3}} \int_\gamma d z\, \frac{z^{5/3}}{1+z^4} \\
&=& \frac{2}{1-e^{2\pi i/3}} 2\pi i \sum\mathrm{Res} \, \frac{z^{5/3}}{1+z^4} \\
&=& \frac{\pi}{\sqrt{3}}
\end{eqnarray*}$$
where the residues are to be taken from the upper half-plane.
The residue of $f(z)$ at $z=z_0$ is just the coefficient of $(z-z_0)^{-1}$ in the Laurent series
and, of course, $z^{5/3}/(1+z^4)$ has a Laurent series about the zeros of $1+z^4$.
(It does not, however, have a Laurent series about $z=0$.)
A simpler solution
Recognize that the integral $I$ is already in the form
$$\int_0^\infty d t\, t^\beta f(t^2).$$
But
$$\begin{eqnarray*}
\int_0^\infty d t\, t^\beta f(t^2)
&=& \frac{1}{1+e^{\beta \pi i}} \int_{-\infty}^\infty d t\, t^\beta f(t^2) \\
&=& \frac{1}{1+e^{\beta \pi i}} \int_\gamma d t\, t^\beta f(t^2) \\
&=& \frac{1}{1+e^{\beta \pi i}} 2\pi i \sum \mathrm{Res} \, t^\beta f(t^2)
\end{eqnarray*}$$
assuming the contributions from the contour around the branch cut at $z=0$ and the contour at infinity vanish.
(Again, we have closed the contour in the upper half-plane and pick up only the residues residing there.)
It is also important that the singularities of $f$ do not lie on the real line.
The contribution near $z=0$ goes like $\epsilon^{1+1/3} = \epsilon^{4/3}$.
On the semicircle at infinity the integral goes like $R^{1+1/3-2} = R^{-2/3}$.
Thus, $\int_{-\infty}^\infty d t\, t^{1/3}/(1+t^2) = \int_\gamma d t\, t^{1/3}/(1+t^2)$.
There is one residue, at $t=i=e^{i\pi/2}$, so we have halved our work.
We find
$$\begin{eqnarray*}
I &=& \frac{1}{1+e^{\pi i/3}} 2\pi i \frac{e^{i\pi/6}}{2i} \\
&=& \frac{\pi}{2\cos \pi/6} \\
&=& \frac{\pi}{\sqrt{3}}.
\end{eqnarray*}$$
It's
$$\int_0^1\frac{\log(1-x+x^2)\log(1+x-x^2)}{x}dx= -2\sum\limits_{k=1}^\infty \frac{(2k-1)!^2}{(4k)!}\sum\limits_{v=0}^{k-1}\frac{1}{k+v}$$
which is $\enspace\approx -0.0848704554500727311… $ .
Already $\enspace\displaystyle -2\sum\limits_{k=1}^{10} \frac{(2k-1)!^2}{(4k)!}\sum\limits_{v=0}^{k-1}\frac{1}{k+v}\enspace$ gives a good approach.
Note: A closed form for such or comparable series is not known to me.
Proof:
$\displaystyle \int_0^1\frac{\log(1-x+x^2)\log(1+x-x^2)}{x}dx=$
$\displaystyle =\int_0^1\lim\limits_{h\to 0}\frac{((1-x+x^2)^h-1)((1+x-x^2)^h-1)}{h^2x}dx$
$\displaystyle =\lim\limits_{h\to 0}\frac{1}{h^2}\int_0^1\frac{((1-x+x^2)^h-1)((1+x-x^2)^h-1)}{x}dx$
$\displaystyle =\lim\limits_{h\to 0}\frac{1}{h^2}\left(\int_0^1\left(\frac{(1-(x-x^2)^2)^h-1}{x}-\frac{(1-x+x^2)^h-1}{x}-\frac{(1-x+x^2)^h-1}{x}\right)dx\right) $
$\displaystyle =\lim\limits_{h\to 0}\frac{1}{h^2}\int_0^1\left(\sum\limits_{k=1}^\infty \binom h k \left(x^{k-1}(-x(1-x)^2)^k -x^{k-1}(-1+x)^k -x^{k-1}(1-x)^k\right) \right) $
$\displaystyle =\lim\limits_{h\to 0}\frac{1}{h^2}\sum\limits_{k=1}^\infty \binom h k \int_0^1\left(x^{k-1}(-x(1-x)^2)^k -x^{k-1}(-1+x)^k -x^{k-1}(1-x)^k\right)dx $
$\displaystyle =\lim\limits_{h\to 0}\frac{1}{h^2}\sum\limits_{k=1}^\infty \binom h k \left((-1)^k\frac{(2k-1)!(2k)!}{(4k)!} -(1+(-1)^k)\frac{(k-1)!k!}{(2k)!}\right) $
$\displaystyle =-\lim\limits_{h\to 0}\frac{1}{h^2}\sum\limits_{k=1}^\infty \left((-1)^{k-1}\binom h k + 2\binom h {2k}\right) \frac{(2k-1)!(2k)!}{(4k)!}$
$\displaystyle =-\sum\limits_{k=1}^\infty \frac{(2k-1)!(2k)!}{(4k)!}\lim\limits_{h\to 0}\frac{1}{h^2}\left((-1)^{k-1}\binom h k + 2\binom h {2k}\right)$
$\displaystyle =-\sum\limits_{k=1}^\infty \frac{(2k-1)!(2k)!}{(4k)!}\frac{H_{2k-1}-H_{k-1}}{k}= -2\sum\limits_{k=1}^\infty \frac{(2k-1)!^2}{(4k)!}\sum\limits_{v=0}^{k-1}\frac{1}{k+v}$
Additional comment:
$$\int_0^1\frac{\log(1-z(x-x^2))\log(1+z(x-x^2))}{x}dx= -2\sum\limits_{k=1}^\infty z^{2k}\frac{(2k-1)!^2}{(4k)!}\sum\limits_{v=0}^{k-1}\frac{1}{k+v}$$
for $\,z\in\mathbb{C}\,$ and $\,|z|\leq 1\,$ .
Best Answer
$$\int_0^1\dfrac{(x-1)e^x}{(x-e^x)^2}dx\\=\int_0^1-\frac{\frac{e^x-xe^x}{e^{2x}}dx}{(\frac{x}{e^x}-1)^2}\\\stackrel{u=\frac{x}{e^x}-1}{=}\int_{-1}^{\frac{1-e}{e}}-\frac{du}{u^2}\\=\left(\frac{e}{1-e}\right)+1=\frac{1}{e-1}$$