$\int_{0}^{19}((x+8)x^{\frac{1}{3}}(19-x)^{\frac{2}{3}})^{-1} \, dx$ with Sausage/Dogbone Contour 
Compute $$\int\limits_{0}^{19}\frac{1}{(x+8)x^{\frac{1}{3}}(19-x)^{\frac{2}{3}}} \, dx$$

Let $f(z) = \frac{1}{(z+8)(z^{\frac{1}{3}})(19-z)^{\frac{2}{3}}}$ where $\sqrt[3]{z}$ has $-\pi \leq \arg z < \pi$ and $\sqrt[3]{(19-z)}$ has $0 \leq \arg(19-z) < 2\pi$. Consider the "sausage contour" $\gamma$:

Let $\gamma_{\varepsilon}$ denote the line above the real axis and $\gamma_{-\varepsilon}$ below, both of length $19$. Let $\gamma_{n}$ and and $\gamma_{m}$ denote the left and right semicircles, both of radius $\varepsilon$. The poles of interest are at $z=0$ and at $z=-19$. We compute by the residue theorem that
$$\oint_{\gamma}f(z) \, dz = 0 $$
I'm not really sure where to go from here. I'm having trouble following the example on Wikipedia that covers this kind of integral because I don't think I understand branch cuts that well. On top of that, that example makes use of the residue at infinity. I'm not sure why that would make it easier. We only briefed talked about it in class and our professor said we didn't need it for any of the problems, this one included.
 A: If one wishes to forgo evaluating the integral of interest by using the residue at infinity, one can proceed as follows.
Since there is only one singularity outside the region bounded by the closed contour $\gamma$, which is a simple pole at $z=-8$, we can deform the contour to be a circle of radius $R>19$.  
Then, with the branch cuts chosen so that $-\pi \le \arg(z)< \pi$ and $0\le \arg(19-z)<2\pi$ we have
$$\begin{align}
\oint_{\gamma}\frac{1}{(z+8)z^{1/3}(19-z)^{2/3}}\,dz&=\oint_{|z|=R}\frac{1}{(z+8)z^{1/3}(19-z)^{2/3}}\,dz\\\\
&-2\pi i\text{Res}\left(\frac{1}{(z+8)z^{1/3}(19-z)^{2/3}}, z=-8\right)\\\\
&=\int_0^{2\pi}\frac{1}{(Re^{i\phi}+8)R^{1/3}e^{i\phi/3}(19-Re^{i\phi})^{2/3}}\,iRe^{i\phi}\,d\phi\\\\
&-2\pi i\left(\frac{1}{(-8)^{1/3}(27)^{2/3}} \right)\\\\
&=\int_0^{2\pi}\frac{1}{(Re^{i\phi}+8)R^{1/3}e^{-i\phi/3}(19-Re^{i\phi})^{2/3}}\,iRe^{i\phi}\,d\phi\\\\
&-2\pi i\,\frac{e^{i\pi/3}}{18} \tag 1
\end{align}$$
Letting $R\to \infty$, we see that the integral on the right-hand side of $(1)$ vanishes. Hence, we find
$$\oint_{\gamma}\frac{1}{(z+8)z^{1/3}(19-z)^{2/3}}\,dz=-2\pi i\,\frac{e^{i\pi/3}}{18} $$
To finish the problem, we note that the contribution from integrations over $\gamma_n$ and $\gamma_m$ vanish as $\epsilon\to 0$.  This leaves
$$(1-e^{-i4\pi/3})\int_0^{19}\frac{1}{(x+8)x^{1/3}(19-x)^{2/3}}\,dx=-2\pi i\,\frac{e^{i\pi/3}}{18}$$ 
whereupon solving for the integral of interest yields
$$\int_0^{19}\frac{1}{(x+8)x^{1/3}(19-x)^{2/3}}\,dx=\frac{\pi}{18\sin(2\pi/3)}=\frac{\pi}{9\sqrt 3}$$
