In response to the comment, I will do my best to attempt to explain how I choose contours for integration.
I first look at the bounds of integration. If it is over $[0, \infty)$ and even, make it over $(-\infty,\infty)$.
Next, I look at how the function behaves around infinity in the top half of the plane. If it decreases fast enough (e.g. $\frac{\exp(ix)}{x^2+1}$), we can integrate with a semicircle contour. If not, find a value $a$ such that when you integrate a rectangle with vertices at $-R, R, R+ia, -R+ia$ (as $R\to\infty$), the vertical sides disappear and the horizontal integrals are equal when multiplied by a constant.
If the function cannot be made even, there is still some hope left to contour integrate. If the function has a branch cut (e.g. $\frac{\sqrt x}{x^2+1}$), try a keyhole contour if the function decays fast enough around $\infty$. Otherwise, try a rectangle.
If the integrand can be simplified as $f(x+ia) = g(x)$, where the integral of $g$ is known or is in the form $A f(x)$, it may be able to be exploited using a rectangular contour. For example, a rectangular contour of infinite width and height $a$ along the real line along with the knowledge that $\int_{-\infty}^\infty e^{-x^2}\, dx =\sqrt{\pi}$ can easily demonstrate that $\int_{-\infty}^\infty e^{-(x+a)^2}\, dx =\sqrt{\pi}$.
Wedge contours can be used in situations like the rectangular contours can, but instead of having $f(x+ia)$ being well-behaved, we have $f(e^{i\theta}x)$ being well behaved. For example, taking again $f(x) = e^{-x^2}$, we see that $f(xe^{i\pi/4}) = e^{-ix^2}$. Thus, taking a wedge contour with $\theta=\pi/4$ we can deduce from the integral $\int_{0}^\infty e^{-x^2}\, dx =\sqrt{\pi}/2$ that $\int_{0}^\infty \sin(x^2)\, dx =\int_{0}^\infty \cos(x^2)\, dx =\sqrt{\pi/2}/2$.
Other contours exist and can be used (e.g. the trapezoid contour for integrating the gaussian integral), though I've found the above contours work for most standard integrals.
If the contour travels through a pole, indent it with a semicircle - with a simple pole, $z_0$, the contributed value from that integral equals $i\theta\operatorname*{Res}f_{z=z_0}$ where $\theta$ equals the angle traversed around the pole.
It is often convenient to change $\sin$ or $\cos$ in the numerator to $e^{ix}$ (which is better behaived for integration around the top half of the plane) and take the real or imaginary part after integration - this can even be done if the other part diverges.
When dealing with exponents, use trigonometric identities to reduce the function into an exponential that decays. For example, it is easier to deal with $\Re \left[\frac{1-e^{2ix}}{2}\right] = \sin^2(x)$ instead of just $\sin^2(x)$.
PRIMER:
The complex logarithm function is a multi-valued function that is defined as
$$\log(z)=\log(|z|)+i\arg(z) \tag1$$
where $\arg(z)$ is the multivalued argument of $z$.
The function $f(z)=z^c$, where $c\in \mathbb{C}$, is defined as
$$f(z)=e^{c\log(z)} \tag2$$
Therefore, $f(z)$ is also multivalued when $c$ is not an integer.
BRANCH POINT
If $z_0$ is branch point of the multivalued function $f(z)$ then there is no open neighborhood $N(z_0)$ of $z_0$ on which $f$ is continuous. Loosely speaking, we cannot encircle $z_0$ without encountering a discontinuity.
We can see from $(1)$ that $z_0=0$ is a branch point of $\log(z)$. Let $z_0=e^{i\theta_0}$ be a point on the unit circle. Then $\log(z_0)=i\theta_0$.
We travel on the unit circle from $z_0$ by increasing $\arg(z)$ from $\theta_0$ to $\theta_0+2\pi$. While we have returned to $z_0$, the value of $\log(z)$ has jumped from $i\theta_0$ to $i(\theta_0+2\pi)$. (Note that we have tacitly cut the plane along the ray $\theta=\theta_0$).
Inasmuch as $(2)$ defines $z^c$, then for non-integer $c$, $z^c$ shares the branch point singularity of $\log(z)$.
To see the reason that $z=\infty$ is also a branch point, we let $w=1/z$. Since $\log(w)$ has a branch point at $w=0$, then $\log(z)=\log(1/w)$ has a branch point at $\infty$.
INTEGRATION OVER THE KEYHOLE CONTOUR
From the previous discussion, we know that $z^{1/3}$ has logarithmic branch points at $z=0$ and $z=\infty$. We choose to cut the plane along the positive real axis.
With this choice of branch cut, if we approach a point on the positive real axis along a contour in the first quadrant, then $\arg(z)$ approaches $0$. If we approach a point on the positive real axis along a contour in the fourth quadrant, then $\arg(z)$ approaches $2\pi$.
Referring to the diagram in the OP, we can formally parameterize the green (red) segments as $z=x \pm i\epsilon$, $x\in [\sqrt{\nu^2-\epsilon^2},\sqrt{R^2-\epsilon^2})$, where $\nu>0$ is the radius of the blue-colored circular arc centered at the origin and $R$ is the radius of the gray-colored circular arc. Then, we have
\begin{align}
\lim_{\epsilon\to 0}\int_{\sqrt{\nu^2-\epsilon^2}}^{\sqrt{R^2-\epsilon^2}}\frac{(x+ i\epsilon)^{1/3}}{(x+ i\epsilon+1)^2}\,dx&=\int_\nu^R \frac{x^{1/3}}{(x+1)^2}\,dx\\\\
\lim_{\epsilon\to 0}\int_{\sqrt{\nu^2-\epsilon^2}}^{\sqrt{R^2-\epsilon^2}}\frac{(x- i\epsilon)^{1/3}}{(x- i\epsilon+1)^2}\,dx&=\int_\nu^R \frac{x^{1/3}e^{i2\pi/3}}{(x+1)^2}\,dx
\end{align}
FINISHING IT UP
It can be shown that as $\nu\to 0$ and $R\to \infty$, the contributions from the integrals around the circular arcs vanish. This leaves
$$\begin{align}
(1-e^{i2\pi/3})\int_0^\infty \frac{x^{1/3}}{(x+1)^2}\,dx&=2\pi i \text{Res}\left(\frac{z^{1/3}}{(z+1)^2},z=-1\right)\\\\
&=2\pi i\lim_{z\to -1}\frac13z^{-2/3}\\\\
&=\frac{2\pi i}3 e^{-i2\pi/3}
\end{align}$$
Solving for
$$\int_0^\infty\frac{x^{1/3}}{(x+1)^2}\,dx=\frac{2\pi}{3\sqrt{3}}$$
Best Answer
Here are some details of the calculation. Let the keyhole contour be oriented counterclockwise and have the slot on the positive real axis. Call the segment above the real axis $\Gamma_1$, the large circle of radius $R$ $\Gamma_2$, the segement below the real axis $\Gamma_3$ and the circle of radius $\epsilon$ around the origin $\Gamma_4.$ We use the branch of the logarithm with the cut along the positive real axis and returning an argument from zero to $2\pi.$ Note that the poles of $$\frac{1}{z^2+2z+2}$$ are at $$(z+1)^2 + 1 = 0$$ or $$\rho_{0,1} = -1 \pm i.$$ Calling the desired integral $I$, we have $$\left(\int_{\Gamma_1} + \int_{\Gamma_2} + \int_{\Gamma_3} + \int_{\Gamma_4}\right) \frac{\log^2 z}{z^2+2z+2} dz \\= 2\pi i \left(\mathrm{Res}\left(\frac{\log^2 z}{z^2+2z+2}; z = -1 + i\right) + \mathrm{Res}\left(\frac{\log^2 z}{z^2+2z+2}; z = -1 - i\right)\right).$$ Now along the large circle we have $$\left|\int_{\Gamma_2} \frac{\log^2 z}{z^2+2z+2} dz\right| \sim 2\pi R \times \frac{\log^2 R}{R^2} \rightarrow 0$$ as $R\rightarrow\infty.$ Along the small circle we get $$\left|\int_{\Gamma_4} \frac{\log^2 z}{z^2+2z+2} dz\right| \sim 2\pi \epsilon \frac{\log^2\epsilon}{2} \rightarrow 0$$ as $\epsilon\rightarrow 0$ by repeated application of L'Hôpital's rule.
Now the residues are easy to compute because the poles are simple and we obtain $$\mathrm{Res}\left(\frac{\log^2 z}{z^2+2z+2}; z = -1 + i\right) = \frac{1}{2i} \log^2(-1+i)$$ and $$\mathrm{Res}\left(\frac{\log^2 z}{z^2+2z+2}; z = -1 - i\right) = - \frac{1}{2i} \log^2(-1-i)$$ As we actually do the computation of these residues we need to be careful to use the same branch of the logarithm as in the integral. A computer algebra system might use a different branch!
For the first residue we get $$\frac{1}{2i} \left(\frac{1}{2}\log 2 + \frac{3}{4} i\pi\right)^2 = \frac{1}{2i} \left(\frac{1}{4}\log^2 2 - \frac{9}{16} \pi^2 + \frac{3}{4}\log 2 \times i\pi\right)$$ and for the second one $$-\frac{1}{2i} \left(\frac{1}{2}\log{2} + \frac{5}{4} i\pi\right)^2 = -\frac{1}{2i} \left(\frac{1}{4}\log^2 2 - \frac{25}{16} \pi^2 + \frac{5}{4}\log 2 \times i\pi\right)$$ Adding these contributions yields $$\frac{1}{2i} \left(\pi^2 - \frac{1}{2}i\pi \log 2\right).$$
Finally we get $$\left(\int_{\Gamma_1} + \int_{\Gamma_3} \right) \frac{\log^2 z}{z^2+2z+2} dz = 2\pi i \times \frac{1}{2i} \left(\pi^2 - \frac{1}{2}i\pi \log 2\right) = \pi^3 - \frac{1}{2}i\pi^2 \log 2$$ Observe that along $\Gamma_3$ the logarithm term produces (actually $x$ would be a better choice of variable here rather than $z$) $$-(\log^2 z + 4\pi i \log z -4\pi^2).$$ The first of these cancels the integral along $\Gamma_1$ and the third is real so that equating imaginary parts we find $$I = \int_{\Gamma_1} \frac{\log z}{z^2+2z+2} dz = - \frac{1}{4\pi} \left(- \frac{1}{2} \pi^2 \log 2\right) = \frac{1}{8} \pi\log 2.$$ We also get the following bonus integral (comparing real parts) $$ \int_{\Gamma_1} \frac{1}{z^2+2z+2} dz = \frac{\pi^3}{4\pi^2} = \frac{\pi}{4}.$$