A Binet-like integral $\int_{0}^{1} \left(\frac{1}{\ln x} + \frac{1}{1-x} -\frac{1}{2} \right) \frac{x^s }{1-x}\mathrm{d}x$ I met this integral

$$
\int_{0}^{1}
\left(\frac{1}{\ln x} + \frac{1}{1-x} -\frac{1}{2} \right) \frac{ \mathrm{d}x}{1-x} \qquad (*)
$$

while evaluating this log-cosine integral. I made several attemps before being successful.
Find a closed form for

$$
I(s): = \int_{0}^{1}
\left(\frac{1}{\ln x} + \frac{1}{1-x} -\frac{1}{2} \right) \frac{x^s }{1-x}\mathrm{d}x, \quad \Re (s)>-1. \qquad (**)
$$

 A: As stated in the comments, by following the Feynman's approach to integrals we have:
$$\begin{eqnarray*} I'(s) &=& \int_{0}^{1}\left(1+\frac{\log x}{1-x}-\frac{\log x}{2}\right)\frac{x^s}{1-x}dx\\ &=& \frac{1}{2}\zeta(2,s+1)+\int_{0}^{1}\left(1+\frac{\log x}{1-x}\right)\frac{x^s}{1-x}dx\\&=&\frac{1}{2}\zeta(2,s+1)-\frac{1}{2s+2}\phantom{}_3 F_2\left(1,1,2;3,s+2;1\right),\end{eqnarray*}$$
but probably there is a nicer form for the second term.
As a matter of fact, by treating the second term this way:
$$\int_{0}^{1}\left(1+\frac{\log x}{1-x}\right)\frac{x^s}{1-x}dx=\lim_{a\to 0^+}\int_{0}^{1}\left(1+\frac{\log x}{1-x}\right)\frac{x^s}{(1-x)^{1-a}}$$
we get:
$$ I'(s)=\frac{1}{2}\psi'(s+1)+s\,\psi'(s)+\psi(s)-\psi(s+1)-1\tag{1}$$
where $\psi$ is the digamma function $\psi(s)=\frac{\Gamma'(s)}{\Gamma(s)}$. By integrating $(1)$ we have:
 $$ I(s) =
 I(1)+\frac{3\gamma+1}{2}-s+\frac{\psi(s+1)}{2}-\log\Gamma(s+1)+s\,\psi(s).\tag{2}$$
All we need to find a closed expression for $I(s)$ is now to evaluate $I(1)$ or $I(0)$, since by $(2)$:
$$ I(0)=I(1)+\gamma-\frac{1}{2}$$
holds. In virtue of $(2)$ we also have:
$$\lim_{s\to +\infty} I(s) = I(1) + \frac{3\gamma-\log(2\pi)}{2},$$
but since
$$ f(x)=\left(\frac{1}{\log x}+\frac{1}{1-x}-\frac{1}{2}\right)\frac{x}{1-x}$$
is a continuous, positive, increasing and bounded function ($f(x)\leq\frac{1}{12}$) on $(0,1)$, as long as $s\to+\infty$ we have $I(s)\leq\frac{1}{12s}$, so $\lim_{s\to +\infty}I(s)=0$. Hence we have:

$$ I(1) = \frac{\log(2\pi)-3\gamma}{2},\qquad  I(0) = \frac{\log(2\pi)-\gamma-1}{2} \tag{3}$$

and:

$$ I(s) =
 \frac{1+\log(2\pi)}{2}-s+\frac{\psi(s+1)}{2}-\log\Gamma(s+1)+s\,\psi(s).\tag{4}$$

As a bonus, by expanding $(4)$ in a neighbourhood of $+\infty$ and exploiting convexity, we can also show that: $$\frac{1}{12s+8}\leq I(s)\leq\frac{1}{12s+6}$$ holds for any $s\in\mathbb{R}^+$.
A: Recall Binet's formula
$$
\log \Gamma(z)= \left( z-\frac{1}{2}\right)\log z - z + \frac{1}{2}\log(2\pi) + 
\int_0^{\infty} \!
\left(\frac{1}{2} - \frac{1}{x} + \frac{1}{e^{x}-1} \right)\frac{e^{-zx}}{x} \mathrm{d}x,\quad \Re z >0
$$ 
which, upon making  $x = - \log v$, can be written as

$$
\log \Gamma(z)= \! \left( z-\frac{1}{2}\right)\log z - z + \frac{1}{2}\log(2\pi) - \!\!
\int_0^{1} \!
\left(\frac{1}{\log v}+\frac{1}{1-v}-\frac{1}{2}\right)\frac{v^{z-1}}{\log v}\mathrm{d}v, \, \Re z >0
$$

and Gauss' formula 

$$
-\psi(z)+\log z = \int_0^{1} \left(\frac{1}{\log v}+\frac{1}{1-v}\right)v^{z-1} \mathrm{d}v.
$$

One may check that
\begin{multline} 
\displaystyle \left(\frac{1}{\log u} + \frac{1}{1-u} - \frac{1}{2} \right) \frac{u^s}{1-u} = u^{s} \frac{d}{du}\left\{u \left(\frac{1}{\log u} + \frac{1}{1-u}\right)\right\} \, \\
+ \left(\displaystyle \frac{1}{\log u} + \frac{1}{1-u} - \frac{1}{2} \right) \frac{u^s}{\log u} - \frac{1}{2} \left(\displaystyle \frac{1}{\log u} + \frac{1}{1-u}\right)u^s. 
\end{multline}
Hence our integral $I(s)$ is the sum of three integrals.
The first,
\begin{multline} 
I_{1}(s) =\displaystyle \left.\left(\frac{1}{\log u} + \frac{1}{1-u} \right)u^{s+1} \right|_{0}^{1} 
- s \int_{0}^{1}\! 
\displaystyle \left(\displaystyle \frac{1}{\log u} + \frac{1}{1-u}\right)u^{s}
\mathrm{d}u,
\end{multline}
using Gauss' formula,
$$
I_{1}(s) = \frac{1}{2} + s \left( \psi(s+1)-\log \left(s+1\right) \right).
$$
The second, applying Binet's formula,
\begin{multline} 
I_{2} (s)= \displaystyle  \int_{0}^{1}\! \left(\displaystyle \frac{1}{\log u} + 
\frac{1}{1-u} -\frac{1}{2} \right) \frac{u^{s}}{\log u} \mathrm{d}u \\
= -\log \Gamma(s+ 1)+ \left(s + 1/2 \right)\log \left(s+ 1 \right) - s - 1 + \frac{1}{2}\ln(2\pi).
\end{multline}
The third, using Gauss' formula once more
\begin{equation} 
I_{3}(s) = - \frac{1}{2} \displaystyle  \int_{0}^{1}\! \left(\displaystyle \frac{1}{\log u} + \frac{1}{1-u}\right)u^{s} \mathrm{d}u  = 
\frac{1}{2}\left( \psi(s+1)-\log \left(s+1\right) \right).
\end{equation}
Consequently, $ I(s)=I_{1}(s)+I_{2}(s)+I_{3}(s)$  is given by

\begin{multline} 
I(s):= \displaystyle \int_0^{1} \left(\frac{1}{\log u}+
\frac{1}{1-u}-\frac{1}{2}\right) \frac{u^s}{1-u} \mathrm{d}u \\
= -\log \Gamma(s+1) + \left( s +\frac{1}{2}\right) \psi(s+1) - s - \frac{1}{2} + \frac{1}{2}\ln(2\pi).
\end{multline}

We have, with $s=0$,

$$
\int_{0}^{1}
\left(\frac{1}{\log x} + \frac{1}{1-x} -\frac{1}{2} \right) \frac{ \mathrm{d}x}{1-x} = -\frac{1}{2}+\frac{1}{2} \ln (2 \pi)-\frac{1}{2} \gamma.
$$

