How to solve this integral equation

Integrating the denominator by parts leads to

\begin{eqnarray} \mathcal D &=& \int_1^{+\infty}\frac{\log x}{(1-x)x}dx=\\ &=& \frac12\left[\frac{\log^2 x}{(1-x)}\right]_1^{+\infty}-\frac12\int_1^{+\infty}\frac{\log^2x}{(1-x)^2}dx=\\ &=&-\frac12\int_1^{+\infty}\frac{\log^2x}{(1-x)^2}dx=\\ &\stackrel{t=\frac1{x}}{=}&-\frac12\int_1^0\frac{\log^2t}{\left(1-\frac1{t}\right)^2}\left(-\frac1{t^2}\right)dt=\\ &=&-\frac12\int_0^1 \frac{\log^2 t}{(1-t)^2}dt = \\ &=&-\frac12 \cdot\mathcal N, \end{eqnarray} where $\mathcal N$ is your numerator. So $$\frac{\mathcal N}{\mathcal D}=-2.$$