I'll give my humble idea to show the integral is $-\dfrac{\pi}{4}$.
With a change of variables ($x=e^u$) we have that
$$\mathcal{I}=\int\limits_0^\infty {\frac{{\log x}}{{{{\left( {1 + {x^2}} \right)}^2}}}dx = } \int\limits_{ - \infty }^\infty {\frac{{u{e^u}}}{{{{\left( {1 + {e^{2u}}} \right)}^2}}}du} $$
We can write this as
$${\mathcal I} = \int\limits_{ - \infty }^\infty {\frac{{u{e^{ - u}}}}{{{{\left( {{e^{ - u}} + {e^u}} \right)}^2}}}du} $$
Putting $u=-v$ we have that
$${\mathcal I} = \int\limits_{ - \infty }^\infty {\frac{{u{e^{ - u}}}}{{{{\left( {{e^{ - u}} + {e^u}} \right)}^2}}}du} = -\int\limits_{ - \infty }^\infty {\frac{{v{e^v}}}{{{{\left( {{e^{ - v}} + {e^v}} \right)}^2}}}dv} $$
This means that
$$2I = 2\int\limits_0^\infty {\frac{{\log x}}{{{{\left( {1 + {x^2}} \right)}^2}}}dx = } \int\limits_{ - \infty }^\infty {\frac{{u\left( {{e^{ - u}} - {e^u}} \right)}}{{{{\left( {{e^u} + {e^{ - u}}} \right)}^2}}}du} $$
We can write this in terms of the hiperbolic functions, to get
$$2I = 2\int\limits_0^\infty {\frac{{\log x}}{{{{\left( {1 + {x^2}} \right)}^2}}}dx = } - \frac{1}{2}\int\limits_{ - \infty }^\infty {\frac{{u\sinh u}}{{\cosh^2 u}}du} $$
Integration by parts gives ($(\operatorname{sech} u)'=-\dfrac{{\sinh u}}{{\cosh^2 u}}$)
$$ - \int\limits_{ - \infty }^\infty {\frac{{\sinh udu}}{{{{\cosh }^2}u}}} = \left[ {u\operatorname{sech} u} \right]_{ - \infty }^\infty - \int\limits_{ - \infty }^\infty {\frac{{du}}{{\cosh u}}} $$
Finally, you can easily check that
$$\int\limits_{ - \infty }^\infty {\frac{{du}}{{\cosh u}}} = \pi $$
and that $u \operatorname{sech} u$ is odd so the first term in the RHS is zero. Thus
$$\eqalign{
& 2I = 2\int\limits_0^\infty {\frac{{\log x}}{{{{\left( {1 + {x^2}} \right)}^2}}}dx = } - \frac{\pi }{2} \cr
& I = \int\limits_0^\infty {\frac{{\log x}}{{{{\left( {1 + {x^2}} \right)}^2}}}dx = } - \frac{\pi }{4} \cr} $$
$|1-e^{i\theta}|=2\sin{\frac{\theta}{2}}$, so we need to evaluate $I=\int_0^{2\pi}\log |1-e^{i\theta}|d\theta$ as then the answer is $2\pi\log R+I$.
Now $\log|1-z|=\Re{\log(1-z)}$ is harmonic inside the unit disc, so by the mean value theorem for harmonic function, $\int_0^{2\pi}\log |1-re^{i\theta}|d\theta=0$ for any $r<1$.
But $\log|1-e^{i\theta}|$ is obviously integrable on the unit circle (since near $0$, $\log 2\sin{\frac{\theta}{2}}-\log \theta$ is continuos and the latter is clearly integrable, while near $2\pi$ we can use $\sin{\frac{\theta}{2}}=\sin{\frac{2\pi-\theta}{2}}$ and the result at zero) and $\log|1-re^{i\theta}| \to \log|1-e^{i\theta}|$ a.e.
But now if say $0 \le \theta \le \frac{\pi}{100}$ or $0 \le 2\pi-\theta \le \frac{\pi}{100}$, by drawing the perpendicular from $1$ to the $\theta$ ray which has length $|\sin \theta|$ it follows from elementary geometry that $|1-re^{i\theta}| \ge |\sin \theta|$, while for the rest $|1-re^{i\theta}| \ge c >0$ and since $|1-re^{i\theta}| \le 2$, we get that $|\log |1-re^{i\theta}|| \le \max {(\log 2, \log^- c-\log |\sin \theta|)}$ and that is integrable on the unit circle as before, hence we can apply the Lebesgue dominated convergence and conclude that $0=\int_0^{2\pi}\log |1-re^{i\theta}|d\theta \to I$, so $I=0$ and the final answer is $2\pi\log R$
As an aside, there is also a classic real variables proof that $I=0$ using the doubling formula for the $\sin$ and various changes of variable while the above can be expressed in terms of contour integrals if one wishes using $f(z)=\frac {\log (1-z)}{z}$ which is analytic inside the unit disc, or treat the original integral using $f(z)=\frac {\log R(1-z)}{z}$ which has $\log R$ as residue at $0$ etc, but of course it may not quite be what OP had in mind.
(Edit) As asked let's quickly use Cauchy rather than Poisson:
$I_1=\int_0^{2\pi}\log (2R\sin{\frac{\theta}{2}})d\theta=\int_0^{2\pi}\log R|1-e^{i\theta}|d\theta= \int_0^{2\pi} \Re {\log R(1-e^{i\theta})}d\theta=\Re {\int_0^{2\pi} \log R(1-e^{i\theta})d\theta}$
with the last equality holding because $d\theta$ is a real (positive) measure.
But now with the usual $e^{i\theta}=z, d\theta=\frac{1}{iz}dz$ we have;
$I_1=\Re {\int_{|z|=1} \frac{\log R(1-z)}{iz}dz}$
Now by Cauchy ${\int_{|z|=1} \frac{\log R(1-rz)}{iz}dz}= 2\pi \log R$ as the residue at zero of the integrand is $\frac{\log R}{i}$, while it is analytic anywhere else on the closed unit disc when $0 < r <1$. So we need to be able to pass to the limit $ r \to 1$ to conclude that the above unit circle integral (and hence its real part) is $2\pi \log R$ and the same argument as above works since the only problem comes from $\log |1-rz|$ near the boundary as everything else is obviously bounded so the same estimates work to show that we can use the Lebesgue dominated convergence and conclude that $I_1= 2\pi \log R$
(for $|z|=1$ we have $|\frac{\log R(1-rz)}{iz}|\le |\log R|+ |\log (1-rz)| \le |\log R|+ |\Re \log (1-rz)|+ |\Im \log (1-rz)| $ and $|\Re \log (1-rz)|=|\log |1-rz||$ as above, while $|\Im \log (1-rz)|=|\arg (1-rz)| \le \frac{\pi}{2}$ since $\Re (1-rz) >0, |z| =1$)
Best Answer
The way a computation of an integral along an unbounded contour reduces to a sum of residues is often left unjustified, as is done in the linked MO post. In fact this requires an analysis of the behavior of the integrand around $z=\infty$; for $z^{-2}\tanh^3 z$, one takes a bounded part of the contour (say, with $|z|<R$), makes it closed (say, the boundary of $[-R,R]+i\pi[-N,N]$ slit along the positive real axis, where $N$ is an integer), and shows that all the "extra" things vanish in the limit (as $N,R\to\infty$). In our case this doesn't work because of the blow-up of $e^{-z}$ as $\Re z\to-\infty$. Still this can be fixed (see the answer by @Svyatoslav).
Another approach (cf.) is as follows. For $a,b\geqslant 0$ and $0<\Re s<1$ we have $$\int_0^\infty x^{s-2}(e^{-ax}-e^{-bx})\,dx=\frac{\Gamma(s)}{1-s}(b^{1-s}-a^{1-s})$$ (say, use $e^{-ax}-e^{-bx}=x\int_a^b e^{-xy}\,dy$ and justify the $dy\,dx\mapsto dx\,dy$), then we find $$\int_0^\infty x^{s-2}\tanh x\,dx=\frac{\Gamma(s)}{1-s}\sum_{n=0}^\infty\big(2(4n+2)^{1-s}-(4n)^{1-s}-(4n+4)^{1-s}\big)$$ (use $\tanh x=(1-e^{-2x})^2\sum_{n=0}^\infty e^{-4nx}$ and DCT for termwise integration). The sum equals $2^{2-s}(1-2^{2-s})\zeta(s-1)$ (compute $\sum_{n=1}^\infty$ for $\Re s>2$, and use analytic continuation; the last function of $s$ is entire). Thus we get (still for $0<\Re s<1$) $$\int_0^\infty x^{s-2}(\tanh x-xe^{-x})\,dx=\Gamma(s)\big(f(s)-1\big),\\f(s):=\frac{2^{2-s}}{1-s}(1-2^{2-s})\zeta(s-1).$$
It just remains to take $s\to 0^+$; then $f(s)\to 1$, hence the limit is $$\int_0^\infty\frac{\tanh x-xe^{-x}}{x^2}\,dx=f'(0)=1-12\zeta'(-1)-\frac73\log2.$$
The expression stated in the OP now follows from the relation between $\zeta'(-1)$ and $\zeta'(2)$ one obtains from the functional equation for $\zeta$, as well as the value of $\zeta'(0)$. Note also that $1-12\zeta'(-1)=12\log A$, where $A$ is the Glaisher–Kinkelin constant.