Two limits involving integrals: $\lim_{\varepsilon\to 0^+}\left(\int_0^{1-\varepsilon}\frac{\ln (1-x)}{x\ln^p x}dx-f_p(\varepsilon)\right)$, $p=1,2$. By applying the Taylor series expansion to $\ln x$, as $x \to 1$, one has the Laurent series expansion, 
$$
\frac1{\ln x}=-\frac1{1-x}+\frac{1}{2}+O\left(1-x\right)
$$ 
then clearly
$$
\begin{align}
&\lim_{\varepsilon \to 0^+} \int_0^{1-\varepsilon} \frac{\ln (1-x)}{x\ln x}\:dx=\infty 
\\\\&\lim_{\varepsilon \to 0^+} \int_0^{1-\varepsilon} \frac{\ln (1-x)}{x\ln^2 x}\:dx=-\infty.
\end{align}
$$
Thus I'm designing the following related limits. 

Question. Find $f_1(\varepsilon)$, $f_2(\varepsilon)$ and find a closed form of $c_1$, $c_2$ such that
  $$
\begin{align}
&\lim_{\varepsilon \to 0^+} \left(\int_0^{1-\varepsilon} \frac{\ln (1-x)}{x\ln x}\:dx-f_1(\varepsilon)\right)=c_1
 \tag1
\\\\&
\lim_{\varepsilon \to 0^+} \left(\int_0^{1-\varepsilon} \frac{\ln (1-x)}{x\ln^2 x}\:dx-f_2(\varepsilon)\right)=c_2. \tag2
\end{align}
$$


Edit. A complete answer is now given below.
 A: 
Claim. As $\epsilon \to 0$,
  $$ \int_{0}^{1-\epsilon} \frac{\log(1-x)}{x \log^2 x} \, \mathrm{d}x = \frac{1+\log\epsilon}{\epsilon} + \frac{\gamma - 1 - \log(2\pi)}{2} + o(1). $$

Step 1. Applying the substitution $t = -\log x$ followed by integration by parts, we have
$$ \int_{0}^{1-\epsilon} \frac{\log(1-x)}{x (-\log x)^{s+1}} \, \mathrm{d}x = \frac{\eta^{-s}}{s} \log \epsilon + \frac{1}{s} \int_{\eta}^{\infty} \frac{t^{-s}}{\mathrm{e}^t - 1} \, \mathrm{d}t, \tag{1} $$
where $\eta = -\log(1-\epsilon)$. Now we focus on the latter integral, and define
$$ I = I(\eta, s) = \int_{\eta}^{\infty} \frac{t^{-s}}{\mathrm{e}^t - 1} \, \mathrm{d}t. $$
Step 2. Next, we give an explicit formula for $I(\eta, s)$ for small $\eta > 0$. (By small, we mean $\eta \in (0, 2\pi)$.) This is easily done by mimicking the derivation of the functional equation for $\zeta$ by contour integration. To this end, we first restrict ourselves to the case $\Re(s) \in (1, 2)$ and consider the contour
$\hspace{5em}$
Call by $C_{\eta}$ the entire contour and by $\Gamma_{\eta}$ the circular arc. Assuming the branch cut of $\log$ as $[0, \infty)$, this can be computed as
\begin{align*}
&I - \mathrm{e}^{-2\pi i s} I + \int_{\Gamma_{\eta}} \frac{z^{-s}}{\mathrm{e}^z - 1} \, \mathrm{d}z \\
&\hspace{3em} = \int_{C_{\eta}} \frac{z^{-s}}{\mathrm{e}^z - 1} \, \mathrm{d}z
= 2\pi i \sum_{n \in \Bbb{Z}\setminus\{0\}} \underset{z=2\pi i n}{\mathrm{Res}} \; \frac{z^{-s}}{\mathrm{e}^z - 1} \\
&\hspace{6em} = 2i \mathrm{e}^{-i\pi s} (2\pi)^{1-s} \cos\left(\frac{\pi s}{2}\right) \zeta(s).
\end{align*}
This gives
$$ I(\eta, s) = \frac{1}{1 - \mathrm{e}^{-2\pi i s}} \int_{-\Gamma_{\eta}} \frac{z^{-s}}{\mathrm{e}^z - 1} \, \mathrm{d}z + \frac{(2\pi)^{1-s}}{2\sin\left(\frac{\pi s}{2}\right)} \zeta(s) $$
Since both sides define entire functions on all of $\Bbb{C}$, this identity extends to all of $s \in \Bbb{C}$. Now we plug the power series $\frac{z}{\mathrm{e}^z - 1} = \sum_{n=0}^{\infty} \frac{B_n}{n!} z^n$ into the contour integral above to get
\begin{align*}
I(\eta, s)
&= \sum_{n=0}^{\infty} \frac{B_n}{n!} \frac{\eta^{n-s}}{s - n} + \frac{(2\pi)^{1-s}}{2\sin\left(\frac{\pi s}{2}\right)} \zeta(s) \tag{2} \\
&= \sum_{n=0}^{\infty} \frac{B_n}{n!} \frac{\eta^{n-s}}{s - n} + \Gamma(1-s)\zeta(1-s) \tag{3}
\end{align*}
Step 3. Now we consider the case of interest, i.e., $s = 1$. Taking limit as $s \to 1$ to $\text{(2)}$, only the first 2 leading terms of the series are significant (as $\eta \to 0$) and we can easily compute $I(\eta, 1)$ as
$$ I(\eta, 1) = \frac{1}{\eta} + \frac{1}{2} \log \eta + \frac{1}{2}(\gamma - \log(2\pi)) + \mathcal{O}(\eta). $$
Therefore the claim follows by plugging this to $\text{(1)}$ and utilizing the asymptotics
$$ \frac{1+\log \epsilon}{\eta} + \frac{1}{2}\log\eta = \frac{1+\log\epsilon}{\epsilon} - \frac{1}{2} + o(1). $$ 

Addendum 1. As a by-product, we can compute @Marco Cantarini's integral:
$$ \int_{0}^{1} \frac{\log(1-x)}{x} \left( \frac{1}{\log^{2} x} - \frac{1}{(1-x)^2} + \frac{1}{1-x}\right) \, \mathrm{d}x = \frac{\gamma + 1 - \log(2\pi)}{2}. $$

Addendum 2. Another by-product is that, for $N \in \Bbb{Z}$ and $N-1 < \Re(s) < N$ we have
$$ \int_{0}^{\infty} x^{s-1} \left( \frac{1}{\mathrm{e}^x - 1} - \sum_{n=-\infty}^{-N} a_n x^{n} \right) \, \mathrm{d}x = \Gamma(s)\zeta(s), $$
where $a_n$ are Laurent coefficients of $\frac{1}{\mathrm{e}^x - 1}$ at $x = 0$:
$$ \frac{1}{\mathrm{e}^x - 1} = \sum_{n=-\infty}^{\infty} a_n x^n = \sum_{n=-1}^{\infty} \frac{B_{n+1}}{(n+1)!} x^n. $$
A: We can try to use your idea for the other integral. Note that $$\frac{\log\left(1-x\right)}{x\log^{2}\left(x\right)}=\frac{\log\left(1-x\right)}{\left(1-x\right)^{2}}+\frac{\log\left(1-x\right)}{x}\left(\frac{1}{\log^{2}\left(x\right)}-\frac{1}{\left(1-x\right)^{2}}+\frac{1}{1-x}\right)
 $$ so we have $$\int_{0}^{1-\epsilon}\frac{\log\left(1-x\right)}{x\log^{2}\left(x\right)}dx=\frac{\log\left(\epsilon\right)+1}{\epsilon}-1
 $$ $$ +\int_{0}^{1}\frac{\log\left(1-x\right)}{x}\left(\frac{1}{\log^{2}\left(x\right)}-\frac{1}{\left(1-x\right)^{2}}+\frac{1}{1-x}\right)dx+O\left(\epsilon\right).\tag{1}
 $$ Let us define $$F\left(s\right)=\int_{0}^{1}x^{s}\left(\frac{1}{\log^{2}\left(x\right)}-\frac{1}{\left(1-x\right)^{2}}+\frac{1}{1-x}\right)dx
 $$ so if we differentiate twice $$F''\left(s\right)=\int_{0}^{1}x^{s}\left(1-\frac{\log^{2}\left(x\right)x}{\left(1-x\right)^{2}}\right)dx=\frac{1}{s+1}-\int_{0}^{1}\frac{\log^{2}\left(x\right)x^{s+1}}{\left(1-x\right)^{2}}dx
 $$ and we have that $$\int_{0}^{1}\frac{\log^{2}\left(x\right)x^{s+1}}{\left(1-x\right)^{2}}dx=\sum_{k\geq1}k\int_{0}^{1}\log^{2}\left(x\right)x^{s+k}dx
 $$ $$ =2\sum_{k\geq1}\frac{k}{\left(s+k+1\right)^{3}}=2\psi^{\left(1\right)}\left(s+2\right)+\left(s+1\right)\psi^{\left(2\right)}\left(s+2\right)$$ so integrating twice and observing that $$F\left(0\right)=\int_{0}^{1}\left(\frac{1}{\log^{2}\left(x\right)}-\frac{1}{\left(1-x\right)^{2}}+\frac{1}{1-x}\right)dx
 $$ $$=\gamma+\int_{0}^{1}\left(\frac{1}{\log^{2}\left(x\right)}-\frac{1}{\left(1-x\right)^{2}}-\frac{1}{\log\left(x\right)}\right)dx=\gamma-\frac{1}{2}
 $$ since $$\int\left(\frac{1}{\log^{2}\left(x\right)}-\frac{1}{\left(1-x\right)^{2}}-\frac{1}{\log\left(x\right)}\right)dx=\frac{1}{x-1}-\frac{x}{\log\left(x\right)}+c
 $$ and $$F'\left(0\right)=\int_{0}^{1}\left(\frac{1}{\log\left(x\right)}-\frac{\log\left(x\right)}{\left(1-x\right)^{2}}+\frac{\log\left(x\right)}{1-x}\right)dx=\gamma-\frac{\pi^{2}}{6}+1
 $$ since $$\int\left(-\frac{1}{1-x}-\frac{\log\left(x\right)}{\left(1-x\right)^{2}}\right)dx=\frac{\log\left(x\right)}{x-1}+\log\left(x\right)+d
 $$ we have $$F\left(s\right)=\left(s+1\right)\left(\log\left(s+1\right)-\psi^{\left(0\right)}\left(s+2\right)\right)+\frac{1}{2}.
 $$ From $(1)$ we get $$\int_{0}^{1}\frac{\log\left(1-x\right)}{x}\left(\frac{1}{\log^{2}\left(x\right)}-\frac{1}{\left(1-x\right)^{2}}+\frac{1}{1-x}\right)dx
 $$ $$=-\sum_{n\geq1}\frac{1}{n}\int_{0}^{1}x^{n-1}\left(\frac{1}{\log^{2}\left(x\right)}-\frac{1}{\left(1-x\right)^{2}}+\frac{1}{1-x}\right)dx
  $$ $$=-\sum_{n\geq1}\frac{n\left(\log\left(n\right)-H_{n}+\gamma\right)+\frac{1}{2}}{n}=\sum_{n\geq1}\left(H_{n}-\log\left(n\right)-\gamma-\frac{1}{2n}\right)$$ and the closed form of the series can be found here, so 

$$\int_{0}^{1-\epsilon}\frac{\log\left(1-x\right)}{x\log^{2}\left(x\right)}dx-\frac{1+\log\left(\epsilon\right)}{\epsilon}\stackrel{\epsilon\rightarrow0^{+}}{\rightarrow}\frac{\gamma-1-\log\left(2\pi\right)}{2}.$$

A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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$\large p = 1$:
\begin{align}
&\int_{0}^{1 - \epsilon}{\ln\pars{1 - x} \over x\ln\pars{x}}\,\dd x =
\int_{0}^{1 - \epsilon}\bracks{%
{\ln\pars{1 - x} \over x\ln\pars{x}} +
{\ln\pars{1 - x} \over x\pars{1 - x}}}\,\dd x +
\int_{0}^{1 - \epsilon}{\ln\pars{1 - x} \over x\pars{1 - x}}\,\dd x
\\[1cm] = &\
\int_{0}^{1}\bracks{%
{\ln\pars{1 - x} \over x\ln\pars{x}} +
{\ln\pars{1 - x} \over x\pars{1 - x}}}\,\dd x 
- \int_{1 - \epsilon}^{1}\bracks{%
{\ln\pars{1 - x} \over x\ln\pars{x}} +
{\ln\pars{1 - x} \over x\pars{1 - x}}}\,\dd x
\\[5mm] - &\
{1 \over 2}\,\ln^{2}\pars{\epsilon} - \,\mrm{Li}_{2}\pars{1 - \epsilon}
\end{align}

$$
\lim_{\epsilon \to 0}\bracks{%
\int_{0}^{1 - \epsilon}{\ln\pars{1 - x} \over x\ln\pars{x}}\,\dd x +
{1 \over 2}\,\ln^{2}\pars{\epsilon}} =
-\,{\pi^{2} \over 6}\ +\ 
\underbrace{\int_{0}^{1}\bracks{%
{\ln\pars{1 - x} \over x\ln\pars{x}} +
{\ln\pars{1 - x} \over x\pars{1 - x}}}\,\dd x}
_{\ds{\approx\ -0.9162}}\ \equiv\ c_{1}
$$


The same procedure can be applied to the case $\ds{p = 2}$.

