Contour integral of $\int_0^\infty \frac{x^{1/3}}{(x+1)(x+2)}dx$ So far I've found that this integral $I$ can be represented with $I' = \oint_C \frac{z^{1/3}}{(z+1)(z+2)}dz$ where $z = re^{i\theta}$ with a branch cut along the x-axis. This gives $$\int_0^\infty \frac{r^{1/3}}{(r+1)(r+2)}dr + \int_\infty^0 \frac{r^{1/3}e^{2/3 \pi i}}{(r+1)(r+2)}dr$$ $$=I-e^{2/3 \pi i}I$$ however I am unsure where to go from here. Any help completing this problem is appreiated!
 A: Let $I$ be the integral given by 
$$I=\int_0^\infty \frac{x^{1/3}}{(x+1)(x+2)}\,dx \tag 1$$

Next, we analyze the closed-contour integral $J$ given by
$$\begin{align}
J&=\oint_C \frac{z^{1/3}}{(z+1)(z+2)}\,dz\\\\
&=\int_0^R \frac{x^{1/3}}{(x+1)(x+2)}\,dx +\int_R^0 \frac{x^{1/3}e^{i2\pi/3}}{(x+1)(x+2)}\,dx\\\\
&+\int_0^{2\pi}\frac{R^{1/3}e^{i\phi/3}}{(Re^{i\phi}+1)(Re^{i\phi}+3)}\,iRe^{i\phi}\,d\phi\tag 2
\end{align}$$
where $C$ is the classical "keyhole" contour.  

For $R>3$, applying the Residue Theorem to $(2)$ reveals 
$$\begin{align}
J&=2\pi i \text{Res}\left(\frac{z^{1/3}}{(z+1)(z+2)}, z=-1,-3 \right)\\\\
&=2\pi i\left(e^{i\pi/3}-2^{1/3}e^{i\pi/3}\right)\\\\
&=2\pi ie^{i\pi/3}(1-2^{1/3})\tag 3
\end{align} $$

Then, letting $R\to \infty$ in $(2)$ and appealing to $(1)$, we obtain
$$\lim_{R\to \infty}J=I(1-e^{i2\pi/3}) \tag 4$$

Setting $(3)$ equal to $(4)$ and solving for $I$ yields

$$I=\frac{2\pi(2^{1/3}-1)}{\sqrt 3}$$

And we are done!
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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It's interesting to show a  $\ds{\bbox[#efe,10px]{'real\ integration'}}$:

\begin{align}
&\int_{0}^{\infty}{x^{1/3} \over \pars{x + 1}\pars{x + 2}}\,\dd x  =
\lim_{\Lambda \to \infty}\sum_{n = 1}^{2}\pars{-1}^{n + 1}
\int_{0}^{\Lambda}{x^{1/3} \over x + n}\,\dd x
\\[5mm] = &\
\lim_{\Lambda \to \infty}\sum_{n = 1}^{2}\pars{-1}^{n + 1}\,n^{1/3}
\int_{0}^{\Lambda/n}{x^{1/3} \over x + 1}\,\dd x
\\[5mm] \stackrel{x\ =\ 1/t - 1}{=}\,\,\, &\
\lim_{\Lambda \to \infty}\sum_{n = 1}^{2}\pars{-1}^{n + 1}\,n^{1/3}
\int_{1}^{n/\pars{\Lambda + n}}t\pars{{1 \over t} - 1}^{1/3}
\pars{-\,{1 \over t^{2}}}\dd t
\\[5mm] = &\
\lim_{\Lambda \to \infty}\sum_{n = 1}^{2}\pars{-1}^{n + 1}\,n^{1/3}
\int_{n/\pars{\Lambda + n}}^{1}t^{-4/3}\,\pars{1 - t}^{1/3}\,\dd t
\\[5mm] = &\
\lim_{\Lambda \to \infty}\sum_{n = 1}^{2}\pars{-1}^{n + 1}\,n^{1/3}\bracks{%
3\pars{n \over \Lambda + n}^{-1/3}\pars{1 - {n \over \Lambda + n}}^{1/3} - \int_{n/\pars{\Lambda + n}}^{1}t^{-1/3}\,\pars{1 - t}^{-2/3}\,\dd t}
\\[5mm] = &\
-\sum_{n = 1}^{2}\pars{-1}^{n + 1}\,n^{1/3}
\int_{0}^{1}x^{-1/3}\pars{1 - x}^{-2/3}\,\dd x =
-\pars{1 - 2^{1/3}}\,{\Gamma\pars{2/3}\Gamma\pars{1/3} \over \Gamma\pars{1}} =
\\[5mm] & =
\pars{2^{1/3} - 1}\,{\pi \over \sin\pars{\pi/3}} =
\bbx{\ds{{2\pars{2^{1/3} - 1}\root{3} \over 3}\,\pi}}
\end{align}
