Show that, $(s+1)\gamma-\int_0^1\left(\frac{1}{\ln(x)}+\frac{1}{1-x}\right)\sum_{n=0}^{s}x^ndx=\sum_{i=1}^{s}H_i-\ln(s+1)!$ harmonic numbers
$H_n=\sum_{i=1}^{n}\frac{1}{i}$
Euler's constant
$\gamma=\lim_{n\to \infty} [H_n-\ln(n)]$
Factorial
$n!=n(n-1)(n-2)\cdots2\cdot1$; valid for all non-negative integers
Show that,
$$(s+1)\gamma-\int_0^1\left(\frac{1}{\ln(x)}+\frac{1}{1-x}\right)\sum_{n=0}^{s}x^ndx=\sum_{i=1}^{s}H_i-\ln(s+1)!$$

Inspiring from the integral of Euler's constant; see Wikipedia
$$\int_0^1\left(\frac{1}{\ln(x)}+\frac{1}{1-x}\right)=\gamma$$
Mathematical experimental
We did trial and error using wolfram integrator and observe the numerical values of the integral and we was able to come up with a closed form for it. 
Unfortunately we are unable to provide a proof for it, can anyone provide us a prove of it?
 A: It is enough to check that:
$$\begin{eqnarray*}\color{blue}{\int_{0}^{1}\left(\frac{1}{\log x}+\frac{1}{1-x}\right)x^k}\,dx &=&\int_{0}^{+\infty}\left(\frac{1}{e^t-1}-\frac{1}{t e^t}\right)e^{-kt}\,dt \tag{1}\\[0.2cm]&=&\gamma+\int_{0}^{+\infty}\frac{e^{-kt}-1}{e^t-1}\,dt-\int_{0}^{+\infty}\frac{e^{-kt}-1}{te^t}\,dt\tag{2}\\[0.2cm]&=&\color{blue}{\gamma-H_k+\log(k+1)}\tag{3}\end{eqnarray*} $$
then sum over $k$. Explanation:


*

*$(1)$ : substitution $x=e^{-t}$;

*$(2)$ : we exploit $\int_{0}^{+\infty}\left(\frac{1}{e^t-1}-\frac{1}{te^t}\right)\,dt=\gamma$ and break the remaining integral in two parts;

*$(3)$ : the first integral can be computed by just expanding the integrand function as a geometric sum and exploiting $\int_{0}^{+\infty}e^{-nt}\,dt=\frac{1}{n}$; the second integral can be computed from Frullani's integral.

A: Here is my two pence worth and is not a complete answer but paves the way I think.
Theorem [Euler 1731]
The limit
$$\gamma = \lim_{n \rightarrow \infty}(H_n-\log n)$$
Is given by the convergent series
$$\gamma = \sum_{n=2}^{\infty}(-1)^{n}\frac{\zeta(n)}{n}$$
We can evaluate this formula using Mercator's expansion
$$\log(1+x) = \sum_{k=1}^{\infty}(-1)^{k}\frac{x^{k}}{k}$$
and in doing so we find
\begin{align}
\log 2 &= 1-\frac{1}{2}\left(\frac{1}{1}\right)^{2}+\frac{1}{3}\left(\frac{1}{1}\right)^{3}-\ldots\\
\log \frac{3}{2} &= \frac{1}{2}-\frac{1}{2}\left(\frac{1}{2}\right)^{2}+\frac{1}{3}\left(\frac{1}{2}\right)^{3}-\ldots\\
\log \frac{4}{3} &= \frac{1}{3}-\frac{1}{2}\left(\frac{1}{3}\right)^{2}+\frac{1}{3}\left(\frac{1}{3}\right)^{3}-\ldots\\
\log \frac{5}{4} &= \frac{1}{4}-\frac{1}{2}\left(\frac{1}{4}\right)^{2}+\frac{1}{3}\left(\frac{1}{4}\right)^{3}-\ldots\\
\end{align}
summing columns of the first $n$ terms gives
$$\log(n+1) = H_n-\frac{1}{2}H_{n, 2}+\frac{1}{3}H_{n, 3}-\ldots$$
or
$$H_n-\log(n+1) = \frac{1}{2}H_{n, 2}-\frac{1}{3}H_{n, 3}+\ldots$$
