Feynman isn't necessarily antangonistic to contour integration. They sometimes work together well.
$$J=\int^\infty_0\frac{\log^4(x)}{1+x^2}dx=I^{(4)}(0)$$
where
$$I(a)=\int^\infty_{0}\frac{x^{a}}{1+x^2}dx$$ with $-1<a<1$.
Take $C$ as a keyhole contour, centered at the origin, avoiding the positive real axis.
Let $f(z)=z^a(1+z^2)^{-1}$ with branch cut on the positive real axis, implying $z^a\equiv \exp(a(\ln|z|+i\arg z))$ where $\arg z\in[0,2\pi)$.
Firstly, by residue theorem,
$$\oint_C f(z)dz=2\pi i\bigg(\operatorname*{Res}_{z=i}f(z)+\operatorname*{Res}_{z=-i}f(z)\bigg)$$
We have
$$\operatorname*{Res}_{z=i}f(z)=\frac{\exp(a(\ln|i|+i\arg i))}{i+i}=\frac{e^{\pi ia/2}}{2i}$$
$$\operatorname*{Res}_{z=-i}f(z)=\frac{\exp(a(\ln|-i|+i\arg -i))}{-i-i}=-\frac{e^{3\pi ia/2}}{2i}$$
Thus,
$$\oint_C f(z)dz=\pi(e^{\pi ia/2}-e^{3\pi ia/2})$$
On the other hand,
$$\oint f(z)dz=K_1+K_2+K_3+K_4$$
where
$$
K_1=\lim_{R\to\infty}\int^{2\pi}_0 f(Re^{it})iRe^{it}dt
=\lim_{R\to\infty}2\pi f(Re^{ic})iRe^{ic}=0 \qquad{c\in[0,2\pi]}$$
$$K_2=\lim_{r\to0^+}\int_{2\pi}^0 f(re^{it})ire^{it}dt
=\lim_{r\to0^+}2\pi f(re^{ic})ire^{ic}=0 \qquad{c\in[0,2\pi]}$$
$$K_3=\int^\infty_0 f(te^{i0})dt=\int^\infty_0\frac{e^{i0}t^a}{t^2+1}dt=I$$
$$K_4=\int_\infty^0 f(te^{i2\pi})dt=-\int^\infty_0\frac{e^{2\pi ia}t^a}{t^2+1}dt=-e^{2\pi ia}I$$
For $K_1,K_2$, please respectively note the asymptotics $f(z)\sim z^{a}$ for small $|z|$ and $f(z)=O(z^{a-2})$ for large $|z|$.
Therefore,
$$I-e^{2\pi ia}I=\pi(e^{\pi ia/2}-e^{3\pi ia/2})$$
$$\implies I=\pi\frac{e^{\pi ia/2}-e^{3\pi ia/2}}{1-e^{2\pi ia}}
=\pi\frac{e^{-\pi ia/2}-e^{\pi ia/2}}{e^{-\pi i a}-e^{\pi ia}}
=\pi\frac{\sin(\pi a/2)}{\sin(\pi a)}
=\frac{\pi}2\sec\left(\frac{\pi a}2\right)
$$
Let $T=\tan(\pi a/2)$, $S=\sec(\pi a/2)$.
$$I^{(4)}(a)=\frac{\pi}2\frac{\pi^4(T^4+18S^2T^2+5S^4)}{16}$$
Hence,
$$J=I^{(4)}(0)=\frac{\pi}2\frac{\pi^4(0+0+5\cdot 1)}{16}=\color{red}{\frac{5\pi^5}{32}}$$
The tedious differentiation is done by calculator. :)
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,
$I = \frac{\pi}{2} 1n(\sqrt{2}+1)$
$\int_1^\infty \frac{arccosh(x)}{x^2+1}dx = \frac{\pi}{2} 1n(\sqrt{2}+1) = \frac{\pi}{2} arcsinh(1)$
Best Answer
Consider for $\xi \lt 0$ the contour integral
$$\oint_C dz \frac{e^{-i \xi z}}{2-2 z-z^2} $$
where $C$ is a semicircle of radius $R$ in the upper half plane, with a small semicircular indent into the upper half-plane of radius $\epsilon$ at each pole $z_{\pm}=-1 \pm \sqrt{3}$. The contour integral is then
$$PV \int_{-R}^R dx \frac{e^{-i \xi x}}{2-2 x-x^2} +i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{e^{-i \xi (z_-+\epsilon e^{i \phi})}}{3-(-\sqrt{3}+\epsilon e^{i \phi})^2}\\ + i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{e^{-i \xi (z_++\epsilon e^{i \phi})}}{3-(\sqrt{3}+\epsilon e^{i \phi})^2}+ i R \int_0^{\pi} d\theta \, e^{i \theta} \frac{e^{-i \xi R e^{i \theta}}}{2-2 R e^{i \theta} - R^2 e^{i 2 \theta}}$$
In the limit as $R \to \infty$, the fourth integral vanishes as $1/R^2$. As $\epsilon \to 0$, the sum of the second and third integrals converge to
$$i \frac{\pi}{2 \sqrt{3}} \left (-e^{-i \xi z_-} + e^{-i \xi z_+} \right ) $$
By Cauchy's theorem, the contour integral is equal to zero. Thus, for $\xi \lt 0$ we have
$$PV \int_{-\infty}^{\infty} dx \frac{e^{-i \xi x}}{2-2 x-x^2} = -\frac{\pi}{\sqrt{3}} e^{i \xi} \sin{\left (\sqrt{3} \xi \right )} $$
For $\xi \gt 0$, one uses a contour in the lower half plane.