How to solve this integral equation While solving a question, I got stuck at

$$\dfrac{\int_{0}^{1}{\dfrac{\ln^2\left(x\right)}{\left(1-x\right)^2}}dx}{\int_{1}^{\infty}{\dfrac{\ln\left(x\right)}{\left(1-x\right)x}}dx}$$

How should I proceed, what I want is to convert $\mathcal{N}$ as $k\mathcal{D}$.
 A: 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 
$$\frac{\mathcal N}{\mathcal D}=-2.$$
A: We have that by parts
$$\int_{1}^{\infty}{\dfrac{\ln\left(x\right)}{\left(1-x\right)x}}dx=\left[\dfrac{\ln^2\left(x\right)}{\left(1-x\right)}\right]_1^\infty-\int_{1}^{\infty}{\frac{\ln x}{x(1-x)}+\dfrac{\ln^2\left(x\right)}{\left(1-x\right)^2}}dx$$
$$\implies \int_{1}^{\infty}{\dfrac{\ln\left(x\right)}{\left(1-x\right)x}}dx=-\frac12\int_{1}^{\infty}{\dfrac{\ln^2\left(x\right)}{\left(1-x\right)^2}}dx$$
and by $y=\frac1x$
$$-\int_{1}^{\infty}{\dfrac{\ln^2\left(x\right)}{\left(1-x\right)^2}}dx=-\int_{0}^{1}{\dfrac{\ln^2\left(y\right)}{\left(1-y\right)^2}}dx$$
