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 i ln(\sqrt{2}+1) $
In conclusion,
$I = \frac{1}{2} 1n(\sqrt{2}+1)$
$\int_1^\infty \frac{arccosh(x)}{x^2+1}dx = \frac{1}{2} 1n(\sqrt{2}+1) = \frac{1}{2} arcsinh(1)$
