Evaluate $\int_{0}^{1}\left [ \frac{1+\sqrt{1-x}}{x} +\frac{2}{\ln\left ( 1-x \right )}\right ]\, \mathrm{d}x$ Evaluate

$$\int_{0}^{1}\left [ \frac{1+\sqrt{1-x}}{x} +\frac{2}{\ln\left ( 1-x \right )}\right ]\, \mathrm{d}x$$

I tried to let $1-x\rightarrow x$ ,and got
$$\int_{0}^{1}\left [ \frac{1+\sqrt{1-x}}{x} +\frac{2}{\ln\left ( 1-x \right )}\right ]\, \mathrm{d}x=\int_{0}^{1}\left ( \frac{1+\sqrt{x}}{1-x}+\frac{2}{\ln x}\right )\, \mathrm{d}x$$
but I have been stuck here for a long time, any idea on how to go futher?
Edit:with the help of Mathematica,I got an answer below

 A: $$\gamma = \sum_{n\geq 1}\left(\frac{1}{n}-\log\frac{n+1}{n}\right) \stackrel{F}{=} \int_{0}^{+\infty}\sum_{n\geq 1}\left(e^{-nx}-\frac{e^{-nx}-e^{-(n+1)x}}{x}\right)\,dx \tag{1}$$
where $F$ stands for Frullani's theorem. That leads to:
$$ \gamma = \int_{0}^{+\infty}\left(\frac{1}{e^x-1}-\frac{1}{x e^x}\right)\,dx  = \int_{0}^{1}\left(\frac{1}{\log u}+\frac{1}{1-u}\right)\,du \tag{2}$$
so the given integral equals:
$$ \int_{0}^{1}\left(\frac{1+\sqrt{u}}{1-u}+\frac{2}{\log u}\right)\,du = 2\gamma+\int_{0}^{1}\frac{-1+\sqrt{u}}{1-u}\,du\tag{3}$$
or:
$$ 2\gamma - \int_{0}^{1}\frac{du}{1+\sqrt{u}} = 2\gamma -\int_{0}^{1}\frac{2v\,dv}{1+v} = \color{red}{2(\gamma+\log 2-1)}\tag{4}$$
as wanted.
A: Let us consider $$I\left(\epsilon\right)=\int_{\epsilon}^{1}\frac{1+\sqrt{1-x}}{x}dx+2\int_{\epsilon}^{1-\epsilon}\frac{1}{\log\left(1-x\right)}dx
 $$ we get $$\int_{\epsilon}^{1}\frac{1+\sqrt{1-x}}{x}dx=-\log\left(\epsilon\right)+2\int_{0}^{\sqrt{1-\epsilon}}\frac{u^{2}}{1-u^{2}}du
 $$ $$=-\log\left(\epsilon\right)-2\sqrt{1-\epsilon}-\log\left(\frac{1-\sqrt{1-\epsilon}}{1+\sqrt{1-\epsilon}}\right)
 $$ and $$2\int_{\epsilon}^{1-\epsilon}\frac{1}{\log\left(1-x\right)}dx=2\int_{\epsilon}^{1-\epsilon}\left(\frac{1}{\log\left(x\right)}+\frac{1}{1-x}\right)dx+2\log\left(\epsilon\right)-2\log\left(1-\epsilon\right)
 $$ then $$I=\int_{0}^{1}\left(\frac{1+\sqrt{1-x}}{x}+\frac{2}{\log\left(1-x\right)}\right)dx=\lim_{\epsilon\rightarrow0^{+}}\left(-2\sqrt{1-\epsilon}-\log\left(\frac{\left(1-\epsilon\right)\left(1-\sqrt{1-\epsilon}\right)}{\left(1+\sqrt{1-\epsilon}\right)\epsilon}\right)+2\int_{\epsilon}^{1-\epsilon}\left(\frac{1}{\log\left(x\right)}+\frac{1}{1-x}\right)dx\right)\tag{1}
 $$ $$=\color{blue}{-2+\log\left(4\right)+2\gamma}.$$ The calculation of the integral in the RHS of $(1)$ is classical and can be found here.
A: 
I have been stuck here for a long time, any idea on how to go further?

Yes. For starters, rewrite $~\dfrac{1+\sqrt x}{1-x}~$ as $~\dfrac1{1-\sqrt x}~,~$ then let $x=t^2,~$ and rewrite $~\dfrac t{1-t}~$ as 
$1-\dfrac1{1-t}.~$ You'll be eventually left with $\displaystyle\int_0^1\left(\frac1{1-x}~+~\frac x{\ln x}\right)~dx,~$ which is much better 
looking. Taking each part individually, and inserting certain helpful parameters, we have 
$\displaystyle\int_0^1\frac{x^\epsilon}{(1-x)^{1-\epsilon}}~dx~=~B(\epsilon,~1+\epsilon),~$ and $~\displaystyle\int_0^1\frac{x}{\ln^{1-\epsilon}x}~dx~=~-\left(-\frac12\right)^\epsilon~\Gamma(\epsilon),~$ ultimately 
leaving us with evaluating $~\displaystyle\lim_{\epsilon\to0^+}~B(\epsilon,~1+\epsilon)~-~\frac{\Gamma(\epsilon)}{2^\epsilon}~=~\gamma~+~\ln2.$ 
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
&\int_{0}^{1}\bracks{{1 + \root{1 - x} \over x} + {2 \over \ln\pars{1 - x}}}
\,\dd x
\,\,\,\stackrel{\root{1 - x}\ \mapsto\ x}{=}\,\,\,
2\int_{0}^{1}{x \over 1 - x}\
\overbrace{\bracks{1 + {1 - x \over \ln\pars{x}}}}
^{\ds{\int_{0}^{1}\pars{1 - x^{t}}\,\dd t}}\ \,\dd x
\\[5mm] = &\
2\int_{0}^{1}\int_{0}^{1}{x - x^{t + 1} \over 1 - x}\,\dd x\,\dd t =
2\int_{0}^{1}\pars{\int_{0}^{1}{1 - x^{t + 1} \over 1 - x}\,\dd x - \int_{0}^{1}\dd x}\dd t
\\[5mm] = &\
2\int_{0}^{1}\braces{\vphantom{\Large A}\bracks{\vphantom{\large A} \Psi\pars{t + 2} + \gamma} - 1}\,\dd t\qquad
\pars{~\Psi:\ Digamma\ Function~}\label{1}\tag{1}
\\[5mm] = &\
2\bracks{\ln\pars{\Gamma\pars{3} \over \Gamma\pars{2}} + \gamma - 1}\qquad
\pars{~\Gamma:\ Gamma\ Function~}
\\[5mm] = &\ \bbx{\ds{2\bracks{\vphantom{\large A}\gamma + \ln\pars{2} - 1}}}
\end{align}

In \eqref{1}, I used the well known identity $\ds{\mathbf{6.3.22}}$ from
  A & S Table.

