Concerning this integral $\int_{0}^{1}\left({1\over \ln(x)}+{1\over 1-x} -{x^s\over 2}\right){\mathrm dx\over 1-x}$ Motivation from this question $(1)$
$$\int_{0}^{1}\left({1\over \ln(x)}+{1\over 1-x} -{1\over 2}\right){x^s\over 1-x}\mathrm dx=F(s)\tag1$$
setting $s=0$ then $F(0)=-{1\over 2}+{1\over 2}\ln(2\pi)-{1\over 2}\gamma$
by a slight variation of $(1)$, we have
$$\int_{0}^{1}\left({1\over \ln(x)}+{1\over 1-x} -{x^s\over 2}\right){\mathrm dx\over 1-x}={1\over 2}H_s-{1\over 2}+{1\over 2}\ln(2\pi)-{1\over 2}\gamma\tag2$$
$H_s$ is the harmonic number, $H_0=0$
How do we go about to prove $(2)?$

Use 
\begin{eqnarray*}
H_v= \int_0^1 \frac{1-x^v}{1-x} dx.
\end{eqnarray*}
We have 
\begin{eqnarray*}
\int_0^1 \left( \frac{1}{\ln x}+ \frac{1}{1-x}-\frac{x^s}{2}\right)\frac{ dx}{1-x} &=& \int_0^1 \left( \frac{1}{\ln x}+ \frac{1}{1-x}-\frac{1}{2} +\frac{1-x^s}{2}\right)\frac{ dx}{1-x} \\
&=& F(0)+ \int_0^1 \frac{1-x^s}{2(1-x)}dx \\
&=&  -\frac{1}{2}+ \frac{1}{2} \ln( 2 \pi) -\frac{1}{2} \gamma + \frac{1}{2} H_s.
\end{eqnarray*}
