The numerical approximation for this intergral is $$\eqalign{J&=-\int_0^\infty\frac{\ln{x}}{x^2+1}\ln\left(1-\frac{\ln{x}}{x^2+1}\right)\,dx\\&\approx0.98484651030802740204425597709883071906753522\dots}$$ Since $\ln x\geq-\ln\left(1-\frac{\ln{x}}{x^2+1}\right)$ holds for all $x\gt 0$, and knowing that $\int_0^\infty\frac{\ln^2(x)}{x^2+1}dx=\frac{\pi^3}{8}$, it can be concluded that $J$ must converge. $J$ bears slight similarity to this question, but the methods used in its answer don't seem readily applicable. I've tried making the substitution $x=e^{-y}$ to get $$\eqalign{J&=\int_{-\infty}^{\infty}\frac{y}{e^y+e^{-y}}\ln\left(1+\frac{y}{e^{-2y}+1}\right)dy\\&=\frac{1}{2}\int_0^\infty y\,\mathrm{sech}(y)\ln\left(\frac{2\cosh(y)+ye^y}{2\cosh(y)-ye^{-y}}\right)dy,}$$ but I'm lost on how to proceed from there. Does $J$ have a closed form? I certainly hope so, even if it involves special functions.
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I've removed the conjecture that was at the start of this question, and I've posed it as its own question here. – teadawg1337 Nov 13 '16 at 16:23