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} $