A partial answer for the moment.
I think such an integral can be attacked with complex analytic techniques, but just after some manipulations. Integrating by parts we have:
$$ I = 2\int_{0}^{+\infty}\frac{t}{1+t^2}(-\log(1-e^{-t}))\,dt =2\sum_{n=1}^{+\infty}\frac{1}{n}\int_{0}^{+\infty}\frac{te^{-nt}}{1+t^2}\,dt\tag{1}$$
and since
$$ \frac{t}{1+t^2} = \int_{0}^{+\infty}e^{-tu}\cos u \,du, \tag{2} $$
it follows that:
$$\begin{eqnarray*} I &=& 2\sum_{n=1}^{+\infty}\frac{1}{n}\int_{0}^{+\infty}\frac{\cos u}{n+u}\,du=\color{red}{2\int_{0}^{+\infty} \frac{H_u}{u} \cos u\,du}\\&=&\color{blue}{2\int_{0}^{+\infty}\frac{dv}{v+1}\sum_{n=1}^{+\infty}\frac{\cos(nv)}{n}},\tag{3}\end{eqnarray*} $$
where $H_u=\gamma+\psi(u+1)$, $\gamma$ is the Euler constant and $\psi=\frac{\Gamma'}{\Gamma}$. In this form, the integral is convergent in virtue of the integral version of the Dirichlet's test: $\color{red}{\frac{H_u}{u}}$ is a smooth function on $\mathbb{R}^+$ decreasing to zero, or $\color{blue}{\sum_{n\geq 1}\frac{\cos(nv)}{n}}$ is a $2\pi$-periodic function, $-\log(2\sin(v/2))$, belonging to $L^1((0,2\pi))$ and having mean zero. We also have:
$$\begin{eqnarray*} I &=& 2\sum_{n=1}^{+\infty}\frac{1}{n}\int_{n}^{+\infty}\frac{\cos u\cos n+\sin u\sin n}{u}\,du\\&=&2\sum_{n=1}^{+\infty}\left(-\frac{\cos n}{n}\operatorname{Ci}(n)+\frac{\sin n}{n}\left(\frac{\pi}{2}-\operatorname{Si}(n)\right)\right)\\&=&\frac{\pi(\pi-1)}{2}-2\sum_{n=1}^{+\infty}\frac{\sin n\operatorname{Si}(n)+\cos n\operatorname{Ci}(n)}{n},\tag{4}\end{eqnarray*}$$
where $\operatorname{Si}(n)=\int_{0}^{n}\frac{\sin z}{z}\,dz$ and $\operatorname{Ci}(n)=-\int_{n}^{+\infty}\frac{\cos z}{z}\,dz$.
Addendum. Binet's second $\log\Gamma$-formula can be seen as a consequence of the Abel-Plana formula and it gives
$$\log\Gamma(z)=\left(z-\frac{1}{2}\right)\log(z)-z+\log\sqrt{2\pi}+\frac{1}{\pi}\int_{0}^{+\infty}\frac{\arctan\frac{t}{2\pi z}}{e^t-1}\,dz.\tag{5}$$
If we consider such identity at $z=\frac{1}{2\pi}$, then apply $\int_{0}^{\frac{1}{2\pi}}\left(\ldots\right)\,dz$ to both sides, we recover a closed form for the similar integral $\int_{0}^{+\infty}\frac{z\log(1+z^2)}{e^z-1}\,dz$. The evaluation at $z=\frac{i}{2\pi}$ leads to a closed form for $\int_{0}^{+\infty}\frac{\text{arctanh}(z)}{e^z-1}\,dz$.
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.
Best Answer
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
$$\eqalign{ & 2I = 2\int\limits_0^\infty {\frac{{\log x}}{{{{\left( {1 + {x^2}} \right)}^2}}}dx = } - \frac{\pi }{2} \cr & I = \int\limits_0^\infty {\frac{{\log x}}{{{{\left( {1 + {x^2}} \right)}^2}}}dx = } - \frac{\pi }{4} \cr} $$