$\int_0^{\infty} \frac{x^{\frac{1}{4}}}{1+x^3} dx = \frac{\pi}{3 \sin\left( \frac{5\pi}{12} \right)}$ I want to evaluate following integral 
\begin{align}
  \int_0^{\infty} \frac{x^{\frac{1}{4}}}{1+x^3} dx = \frac{\pi}{3 \sin\left( \frac{5\pi}{12} \right)}
\end{align}
Simple try on this integral is using branch cut and apply residue theorem. 
Usual procedure gives  for $0 < \alpha < 1$,  with $Q(x)$ deg $n$ and $P(x)$ deg $m$, for $x>0$, $Q(x) \neq 0$
\begin{align}
\int_0^{\infty} \frac{x^\alpha P(x)}{Q(x)} dx = \frac{2\pi i}{1- e^{i\alpha 2 \pi}} \sum_j Res[\frac{z^\alpha P(z)}{Q(z)} , z_j]
\end{align}
where $z_j$ are poles which does not make $\frac{P}{Q}$ be zero.
This formula comes from Mathews and Howell's complex analysis textbook. 
And this is nothing but applying branch cut to make $x^{\frac{1}{4}}$ singled valued function. I think this formula works for above improper integral but results seems different. 
Apply $\alpha=\frac{1}{4}$ and take poles $z_0=-1$, $z_1 = e^{\frac{i \pi}{3}}$
, $z_2 = e^{\frac{i5 \pi}{3}}$, i got different things. 
Am i doing right? 

\begin{align}
\frac{2\pi i}{1-i}\frac{1}{3} \left( e^{\frac{1}{4} \pi i} + e^{-\frac{7}{12}\pi i}    + e^{-\frac{35}{12} \pi i}\right) 
\end{align}
 A: You may simply remove the branch cut by setting $x=z^4$:
$$ I = 4 \int_{0}^{+\infty}\frac{z^4\,dz}{1+z^{12}} = 2\int_{-\infty}^{+\infty}\frac{z^4}{1+z^{12}} \tag{1}$$
and by evaluating the residues at the roots of $1+z^{12}$ in the upper half-plane,
$$ I = \frac{2\pi}{3\sqrt{2+\sqrt{3}}}=\frac{\pi}{3}\left(\sqrt{6}-\sqrt{2}\right)\tag{2} $$
follows. By setting $\frac{1}{1+z^{12}}=u$, the integral $(1)$ can also be evaluated through Euler's Beta function and the reflection formula for the $\Gamma$ function, since:
$$ I = \frac{1}{3}\int_{0}^{1}u^{-5/12}(1-u)^{-7/12}\,du = \frac{1}{3}\,\Gamma\left(\frac{5}{12}\right)\,\Gamma\left(\frac{7}{12}\right)=\frac{\pi}{3\sin\frac{5\pi}{12}}.\tag{3}$$
A: Letting $y=\frac{1}{1+x^3}$ yields
$$
\begin{aligned}
I &=\int_{1}^{0} y\left(\frac{1}{y}-1\right)^{\frac{1}{12}} \frac{1}{3}\left(\frac{y}{1-y}\right)^{\frac{2}{3}} \frac{d y}{-y^{2}} \\
&=\frac{1}{3} \int_{0}^{1} y^{\frac{1}{12}-1}(1-y)^{\frac{5}{12}-1} d y\\&= \frac{1}{3} B\left(\frac{7}{12}, \frac{5}{12}\right)
\end{aligned}
$$
Using the property of Beta function$$
B(x, 1-x)=\pi \csc (\pi x), \quad \textrm{ where } x \notin \mathbb{Z},
$$
we can now conclude that
$$
\boxed{I=\frac{\pi}{3} \csc \frac{5 \pi}{12}}
$$
