Difficulty in evaluating $\int_{0}^{1}\frac{1-x(1-\ln x)}{x\ln^2 x}\cdot \ln(1-x)dx$ I would like to compute
$$\large \int_{0}^{1}\frac{1-x(1-\ln x)}{x\ln^2 x}\cdot \ln(1-x)\mathrm dx$$
making a substitution
$\large x=e^{-u}$
$$I=\int_{0}^{\infty}\frac{1-(1+u)e^{-u}}{u^2}\cdot \ln(1-e^{-u})\mathrm du$$
but find this $I$ impossible to transform it into a simple form.

I would like a hint please.

 A: Denote, for $a>0$, 
$$f(a) = \int_0^\infty {\frac{{1 - (1 + x){e^{ - x}}}}{{{x^2}}}{e^{ - ax}}dx}$$
Then
$$f''(a)=\int_0^\infty {\left[ {1 - (1 + x){e^{ - x}}} \right]{e^{ - ax}}dx} = \frac{1}{{a{{(1 + a)}^2}}}$$
Hence
$$f'(a)=\frac{1}{{1 + a}} + \ln \frac{a}{{1 + a}} + {C_1}$$ for some constant $C_1$, the fact that $f'(a) \to 0$ as $a\to \infty$ gives $C_1=0$.
Integrate again gives
$$f(a) = a\ln \left( {\frac{a}{{1 + a}}} \right) + C_0$$
the fact that $f(a) \to 0$ as $a\to \infty$ gives $C_0=1$.

$$\begin{aligned}
\int_0^\infty {\frac{{1 - (1 + x){e^{ - x}}}}{{{x^2}}}\ln (1 - {e^{ - x}})dx} &= 
- \sum\limits_{n = 1}^\infty \int_0^\infty{\frac{{1 - (1 + x){e^{ - x}}}}{{n{x^2}}}{e^{ - nx}}}dx \\ &= 
-\lim_{N\to\infty} \sum\limits_{n = 1}^N {\frac{1}{n}\left[ {n\ln \left( {\frac{n}{{1 + n}}} \right) + 1} \right]} \\ &= \lim_{N\to\infty} \left[ \ln (1 + N) - {H_N} \right] = - \gamma 
\end{aligned}$$
the exchange of sum and integral is permitted by dominated convergence theorem, note that the finite sum telescopes.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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\begin{align}
&\bbox[10px,#ffd]{\int_{0}^{1}{1 - x\bracks{1 - \ln\pars{x}} \over x\ln^{2}\pars{x}}\,\ln\pars{1 - x}\,\dd x}
\\[5mm] = &\
\int_{0}^{1}\overbrace{\pars{\int_{0}^{1}t\, x^{t}\,\dd t}}
^{\ds{{1 - x\bracks{1 - \ln\pars{x}} \over \ln^{2}\pars{x}}}}
{\ln\pars{1 - x} \over x}\,\dd x
\\[5mm] = &\
\int_{0}^{1}t\
\underbrace{\int_{0}^{1}x^{t - 1}\ln\pars{1 - x}\,\dd x}
_{\ds{-\,{\Psi\pars{t + 1} + \gamma \over t}}}\,\,\,\dd t =
\left.\vphantom{\LARGE A}-\ln\pars{\Gamma\pars{t + 1}} - \gamma\,t\,\right\vert_{\ 0}^{\ 1}
\\[5mm] = &\ \bbx{-\,\gamma} \\&
\end{align}

Note that
\begin{align}
&\bbox[10px,#ffd]{\int_{0}^{1}x^{t - 1}\ln\pars{1 - x}\,\dd x} =
\left.\partiald{}{\mu}\int_{0}^{1}x^{t - 1}\pars{1 - x}^{\mu}\,\dd x
\,\right\vert_{\ \mu\ =\ 0}
\\[5mm] = &\
\left.\partiald{}{\mu}{\Gamma\pars{t}\Gamma\pars{\mu + 1} \over
\Gamma\pars{t + \mu + 1}}
\,\right\vert_{\ \mu\ =\ 0}
\\[5mm] = &\
\Gamma\pars{t}\left.\partiald{}{\mu}{1 - \gamma\mu \over
\Gamma\pars{1 + t} + \Gamma\pars{1 + t}\Psi\pars{1 + t}\mu}
\,\right\vert_{\ \mu\ =\ 0}
\\[5mm] = &\
{1 \over t}\left.\partiald{}{\mu}\pars{1 - \gamma\mu}
\bracks{1 - \Psi\pars{1 + t}\mu}
\,\right\vert_{\ \mu\ =\ 0}
\\[5mm] = &\
{1 \over t}\left.\partiald{}{\mu}\pars{1 - \gamma\mu
-\Psi\pars{1 + t}\mu}\,\right\vert_{\ \mu\ =\ 0}
\\[5mm] = &\
-\,{\Psi\pars{1 + t} + \gamma \over t}
\end{align}
