It would be the situation in B: you would deform around the pole. It works as follows.
The inverse Laplace transform is given by Cauchy's theorem. I present the parametrization of each piece of the contour, assuming that the radius of the semicircular detour about the pole $z=-1$ and the branch point $z=0$ is $\epsilon$:
$$\int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{\sqrt{s} (1+s)} + e^{i \pi} \int_{\infty}^{1+\epsilon} dx \frac{e^{-t x}}{e^{i \pi/2} \sqrt{x} (1-x)}+i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{e^{t (-1+ \epsilon e^{i \phi})}}{\sqrt{e^{i \pi}+\epsilon e^{i \phi}} (\epsilon e^{i \phi})}\\+ e^{i \pi} \int_{1-\epsilon}^{\epsilon} dx \frac{e^{-t x}}{e^{i \pi/2} \sqrt{x} (1-x)}+i \epsilon \int_{\pi}^{-\pi} d\phi \, e^{i \phi} \frac{e^{t \epsilon e^{i \phi}}}{\sqrt{\epsilon e^{i \phi}} (1+\epsilon e^{i \phi})} +e^{-i \pi} \int_{\epsilon}^{1-\epsilon} dx \frac{e^{-t x}}{e^{-i \pi/2} \sqrt{x} (1-x)}\\+ i \epsilon \int_{2 \pi}^{\pi} d\phi \, e^{i \phi} \frac{e^{t (-1+ \epsilon e^{i \phi})}}{\sqrt{e^{-i \pi}+\epsilon e^{i \phi}} (\epsilon e^{i \phi})}+ e^{-i \pi} \int_{1+\epsilon}^{\infty} dx \frac{e^{-t x}}{e^{-i \pi/2} \sqrt{x} (1-x)} = 0$$
Note that the integrals about the semicircular detours above and below the axis (the 3rd and the 7th integrals) cancel. In the limit as $\epsilon \to 0$, the integral about the branch point (the 5th integral) also vanishes. We are then left with, as $\epsilon \to 0$,
$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{\sqrt{s} (1+s)} + \frac1{2 \pi} PV \int_{\infty}^0 dx \frac{e^{-t x}}{\sqrt{x} (1-x)} - \frac1{2 \pi} PV \int_0^{\infty} dx \frac{e^{-t x}}{\sqrt{x} (1-x)} = 0$$
where $PV$ denotes the Cauchy principal value of the integral. Thus, the ILT is given by (subbing $x=u^2$)
$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{\sqrt{s} (1+s)} = \frac1{\pi} PV \int_{-\infty}^{\infty} du \, \frac{e^{-t u^2}}{1-u^2} $$
To evaluate the integral, we rewrite as
$$e^{-t} PV \int_{-\infty}^{\infty} du \, \frac{e^{t (1- u^2)}}{1-u^2} = e^{-t} I(t)$$
where
$$I'(t) = e^{t} PV \int_{-\infty}^{\infty} du \, e^{-t u^2} = \sqrt{\pi} t^{-1/2} e^{t} $$
and $I(0) = 0$. Thus,
$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{\sqrt{s} (1+s)} = e^{-t}\frac1{\pi} \sqrt{\pi} \int_0^t dt' \, t'^{-1/2} e^{t'} = e^{-t} \frac{2}{\sqrt{\pi}} \int_0^{\sqrt{t}} dv \, e^{v^2} $$
or, finally,
$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \frac{e^{s t}}{\sqrt{s} (1+s)} = e^{-t} \operatorname{erfi}{\left (\sqrt{t} \right )} $$
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
I figured it out, this is a really cool problem that requires you to be very careful with the integration paths along the branch cuts.
Firstly, take the branch cut along the principle branch, this would be $(1,\infty]$. Now we consider the contour integral of the following function
$\oint_C \frac{arccosh^2(z)}{z^2+1}dz$
The Paths of C are broken up in the following way
$\oint_C = \int_{C_{1}}+\int_{C_{2}}+\int_{C_{3}}+\int_{C_{4}}+\int_{C_{5}}+\int_{C_{6}}+\int_{C_{\Gamma}}+\int_{C_{\gamma}}$
The integral around $C_{\Gamma}$ is the usual large circle $z = Re^{i\theta}$.
The integral around $C_{\gamma}$ is the indented circle $z = 1+\epsilon e^{i\theta}$, where we indent around $z = 1$.
You can use the fact that $|zf(z)| \rightarrow 0$ as $z \rightarrow \infty$, and that $(z-1)f(z) \rightarrow 0$ as $z \rightarrow 1$, to estimate both of those integrals to be 0.
So all we need to do is consider each path very carefully, as the arccosh(z) is multi-valued across the branch cut. I've taken $\epsilon \rightarrow 0$ as a consequence of the integral along the indented circle to be zero. The reason I broke up $C_{2}$ and $C_{3}$ is because this integral transitions from the negative real axis to the positive real axis, so our choice of parametrization has to reflect this. The same thing applies for $C_{4}$ and $C_{5}$.
$C_{1}: z = -x$ , $x \in (R,1]$
$C_{2}: z = -x$ , $x \in [1,0]$
$C_{3}: z = x$ , $x \in [0,1]$
$C_{4}: z = x$ , $x \in [1,0]$
$C_{5}: z = -x$ , $x \in [0,1]$
$C_{6}: z = -x$ , $x \in [1,R)$
Now Lets look at integrals $C_{2}$ through $C_{5}$. This specifically concerns the branch cut $z \in [-1,1]$ and we must be careful of how we define the $arccosh(z)$ around each path. Generally speaking
$arccosh(z) = \pm ln(i\sqrt{1-z^{2}}+z)$, $z \in [-1,1]$
For $C_{2}$ and $C_{3}$ use the positive branch, for $C_{4}$ and $C_{5}$ use the negative branch.
$\int_{C_{2}}\frac{arccosh^2(z)}{z^2+1}dz+\int_{C_{3}}\frac{arccosh^2(z)}{z^2+1}dz+\int_{C_{4}}\frac{arccosh^2(z)}{z^2+1}dz+\int_{C_{5}}\frac{arccosh^2(z)}{z^2+1}dz$.
To be continued I need sleep!
Edit: It was a long nap!!
$\int_{C_{2}}\frac{(ln(i\sqrt{1-z^{2}}+z))^2}{z^2+1}dz+\int_{C_{3}}\frac{(ln(i\sqrt{1-z^{2}}+z))^2}{z^2+1}dz+\int_{C_{4}}\frac{(-ln(i\sqrt{1-z^{2}}+z))^2}{z^2+1}dz+\int_{C_{5}}\frac{(-ln(i\sqrt{1-z^{2}}+z))^2}{z^2+1}dz$ =
$\int_1^0\frac{ln^2(i\sqrt{1-x^{2}}-x)}{x^2+1}(-dx)+\int_0^1\frac{ln^2(i\sqrt{1-x^{2}}+x)}{x^2+1}dx+\int_1^0\frac{ln^2(i\sqrt{1-x^{2}}+x)}{x^2+1}dx+\int_0^1\frac{ln^2(i\sqrt{1-z^{2}}-x)}{x^2+1}(-dx)$ =
$\int_0^1\frac{ln^2(i\sqrt{1-x^{2}}-x)}{x^2+1}dx+\int_0^1\frac{ln^2(i\sqrt{1-x^{2}}+x)}{x^2+1}dx-\int_0^1\frac{ln^2(i\sqrt{1-x^{2}}+x)}{x^2+1}dx-\int_0^1\frac{ln^2(i\sqrt{1-x^{2}}-x)}{x^2+1}dx$
But all of these terms cancel and the contribution of these integrals is zero. So now we consider the integrals around $C_{1}$ and $C_{6}$. The $arccosh(z)$ has the following values on these integrals.
$arccosh(z) = \pm i\pi+ln(\sqrt{z^{2}-1}-z)$, $z \in (-\infty,-1]$
For $C_{1}$ use the positive branch, for $C_{6}$ use the negative branch.
$\int_{C_{1}}\frac{arccosh^2(z)}{z^2+1}dz+\int_{C_{6}}\frac{arccosh^2(z)}{z^2+1}dz$ =
$\int_{C_{1}}\frac{(i\pi+ln(\sqrt{z^{2}-1}-z))^2}{z^2+1}dz+\int_{C_{6}}\frac{(-i\pi+ln(\sqrt{z^{2}-1}-z))^2}{z^2+1}dz$ =
$\int_R^1\frac{(i\pi+ln(\sqrt{x^{2}-1}+x))^2}{x^2+1}(-dx)+\int_1^R\frac{(-i\pi+ln(\sqrt{x^{2}-1}+x))^2}{x^2+1}(-dx)$ =
$\int_1^R\frac{-\pi^{2}+2\pi iln(\sqrt{x^{2}-1}+x)+ln^2(\sqrt{x^{2}-1}+x)}{x^2+1}dx+\int_1^R\frac{\pi^{2}+2\pi iln(\sqrt{x^{2}-1}+x)-ln^2(\sqrt{x^{2}-1}+x)}{x^2+1}dx$ =
$\int_1^R\frac{-\pi^{2}+2\pi iln(\sqrt{x^{2}-1}+x)+ln^2(\sqrt{x^{2}-1}+x)+\pi^{2}+2\pi iln(\sqrt{x^{2}-1}+x)-ln^2(\sqrt{x^{2}-1}+x)}{x^2+1}dx$
Canceling the appropriate terms, and letting $R \rightarrow \infty$
$4 \pi i \int_1^\infty \frac{ln(\sqrt{x^2-1}+x)}{x^2+1}dx$ = $4 \pi i \int_1^\infty \frac{arccosh(x)}{x^2+1}dx$ = $4 \pi i I$
Where I is the integral we are looking for. Now by construction this is equal to
$\oint_C \frac{arccosh^2(z)}{z^2+1}dz$
And this is equal to the residues that are collected at $z = \pm i$, multiplied by a factor of $2 \pi i$. A quick computation shows,
$\oint_C \frac{arccosh^2(z)}{z^2+1}dz = 4 \pi i I = 2 \pi^2 i ln(\sqrt{2}+1) $
In conclusion,
$I = \frac{\pi}{2} 1n(\sqrt{2}+1)$
$\int_1^\infty \frac{arccosh(x)}{x^2+1}dx = \frac{\pi}{2} 1n(\sqrt{2}+1) = \frac{\pi}{2} arcsinh(1)$