Evaluate $\text{k}$ from the given equation 
If
  $$ \int_{0}^{\infty} \left(\dfrac{\ln x}{1-x}\right)^{2} \mathrm{d}x + \text{k} \times \int_{0}^{1} \dfrac{\ln (1-x)}{x} \mathrm{d}x =0$$  
then find the value of $\text{k}$   

My Approach : 
Let  
$\text{I}= \displaystyle \int_{0}^{\infty} \left(\dfrac{\ln x}{1-x}\right)^{2} \mathrm{d}x + \displaystyle \text{k} \times \int_{0}^{1} \dfrac{\ln (1-x)}{x}\mathrm{d}x$  
$=\displaystyle \int_{0}^{1} \left(\dfrac{\ln x}{1-x}\right)^{2} \mathrm{d}x + \int_{1}^{\infty} \left(\dfrac{\ln x}{1-x}\right)^{2} \mathrm{d}x + \displaystyle \text{k} \times \int_{0}^{1} \dfrac{\ln (1-x)}{x}\mathrm{d}x$
Now, for the second integral, let $x=\dfrac{1}{t}$,  
$\implies \text{I}= \displaystyle \int_{0}^{1} \left(\dfrac{\ln x}{1-x}\right)^{2} \mathrm{d}x + \displaystyle \int_{0}^{1} \left(\dfrac{\ln t}{1-t}\right)^{2} \mathrm{d}t + \displaystyle \text{k} \times \int_{0}^{1} \dfrac{\ln (1-x)}{x}\mathrm{d}x$  
$= 2\displaystyle \int_{0}^{1} \left(\dfrac{\ln x}{1-x}\right)^{2} \mathrm{d}x + \displaystyle \text{k} \times \int_{0}^{1} \dfrac{\ln (1-x)}{x}\mathrm{d}x$  
$=2\displaystyle \int_{0}^{1} \left(\dfrac{\ln x}{1-x}\right)^{2} \mathrm{d}x + \displaystyle \text{k} \times \int_{0}^{1} \dfrac{\ln x}{1-x}\mathrm{d}x$  
However, I can't seem to think of a way to simplify it further and find the value of $\text{k}$.  
Any help will be appreciated.
Thanks in advance.
 A: You can integrate by parts. A primitive to $1/(1-x)^2$ is $1/(1-x)-1=x/(1-x)$. The derivative of $(\ln x)^2$ is $2(\ln x)/x$. Thus
$$
\begin{aligned}
\int_0^1 \frac{(\ln x)^2}{(1-x)^2}\,dx &= \Bigl[\frac{x}{1-x}(\ln x)^2\Bigr]_0^1-\int_0^1 \frac{x}{1-x}\frac{2\ln x}{x}\,dx\\
&=-2\int_0^1\frac{\ln x}{1-x}\,dx.
\end{aligned}
$$
A: The crux to completing your work is to prove the following

$$\int_0^1 \frac{\log^2(1-x)}{x^2} \,\mathrm{d}x
= -2 \int_0^1 \frac{\log(1-x)}{x}\,\mathrm{d}x$$

You are close though, let me show you how to finish your work. I really liked this problem. Summarizing your work, you have proven that 
$$
\int_0^\infty \left(\frac{\log x}{1-x}\right)^2 \,\mathrm{d}x
= 2\int_0^1 \left(\frac{\log x}{1-x}\right)^2 \,\mathrm{d}x
= 2\int_0^1 \frac{\log^2(1-x)}{x^2} \,\mathrm{d}x
$$
Where the substitution $u \mapsto 1-x$ was used in the last equality.
The idea to finish this problem is to use integration by parts with 
$$
\begin{align*}
u & \ = \hspace{0.7cm} \log^2(1-x) \hspace{1.4cm} \mathrm{d}v \ = \ \frac{1}{u^2}\mathrm{d}x\\
\mathrm{d}u & \ =  - 2\,\frac{\log^{\phantom{2}}(1-x)}{1-x}\,\mathrm{d}x \qquad v \ = \ - \frac{1}{x}
\end{align*}
$$
However using integration by parts we have to be extremely gentle. For reasons that shall become apparent later I replace the upper limit with $\varepsilon$. Then at the end we will let $\varepsilon \to 1$.
$$
\begin{align*}
\int_0^\varepsilon \frac{\log^2(1-x)}{x^2} \,\mathrm{d}x
& = \left[ - \frac{\log^2(1-x)}{x} \right]_0^\varepsilon - 2\int_0^\varepsilon \frac{\log(1-x)}{x(1-x)}\,\mathrm{d}x \\
& = -\frac{\log^2(1-\varepsilon)}{\varepsilon}
- 2 \int_0^\varepsilon \log (1-x)\left(\frac{1}{x}+\frac{1}{1-x}\right)\,\mathrm{d}x \\
& = -\frac{\log^2(1-\varepsilon)}{\varepsilon} - 2 \int_0^\varepsilon \frac{\log(1-x)}{x}\,\mathrm{d}x - 2\left[ -\frac{\log^2(1-x)}{2}\right]_0^\varepsilon \\
& = -2 \int_0^1 \frac{\log(1-x)}{x}\,\mathrm{d}x
\end{align*}
$$
Now the rest is elementary
$$
\int_0^\infty \left(\frac{\log x}{1-x}\right)^2 \,\mathrm{d}x
= 2\int_0^1 \frac{\log^2(1-x)}{x^2} \,\mathrm{d}x
= -4 \int_0^1 \frac{\log(1-x)}{x}\,\mathrm{d}x
$$
As wanted

To prove that 
$$
\lim_{\varepsilon \to 1} \left(\log^2(1-\varepsilon) - \frac{\log^2(1-\varepsilon)}{\varepsilon}\right) = 0
$$
Can be proven for example by a clever taylor expansion or plain old l'hoptials rule
$$
\lim_{\varepsilon \to 1} \frac{\log^2(1-\varepsilon)}{\varepsilon/(\varepsilon-1)} \left[ \frac{0}{0}\right]
= \lim_{\varepsilon \to 1} -2(1-\varepsilon)\log(1-\varepsilon)
= -2 \lim_{x \to 0} x \log x
= 0
$$
Where the last limit is know (again l'hôpital or series expansion).
