How do we prove that $\int_{0}^{1}x^n\left({1\over \ln{x}}+{1\over 1-x}\right)dx={\gamma+\ln(1+n)-H_n}$ How do we prove that
n $\in$ $\Re$

$$\int_{0}^{1}x^n\left({1\over \ln{x}}+{1\over 1-x}\right)dx=\color{red}{\gamma+\ln(1+n)-H_n}.\tag1$$

$$J=\int_{0}^{1}{x^n\over 1-x}dx=\sum_{k=0}^{\infty}(-1)^k\int_{0}^{1}x^{n+k}dx=\sum_{k=0}^{\infty}{(-1)^k\over n+k+1}\tag2$$
This integral seem difficult to evaluate, help needed. Thank!
$$\int_{0}^{1}{x^n\over \ln{x}}dx\tag3$$
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\begin{align}
&\color{#f00}{%
\int_{0}^{1}x^{n}\bracks{{1 \over \ln\pars{x}} + {1 \over 1 - x}}\,\dd x}
\\[3mm]\ =\ &\
\underbrace{\int_{0}^{1}\bracks{{1 \over \ln\pars{x}} + {1 \over 1 - x}}\,\dd x}
_{\ds{\mathrm{1.} = \gamma}}\ -\
\underbrace{\int_{0}^{1}{1 - x^{n} \over \ln\pars{x}}\,\dd x}
_{\ds{\mathrm{2.} = -\ln\pars{n + 1}}}\ -\
\underbrace{\int_{0}^{1}{1 - x^{n} \over 1 - x}\,\dd x}
_{\ds{\mathrm{3.} =H_{n}}}\,,\quad\pars{~\mbox{see below}~}
\\[3mm] = &\
\color{#f00}{\gamma + \ln\pars{n + 1} - H_{n}}
\end{align}


Why ?.



*

*See the proof as Eq. $\pars{2}$ in one of my yesterday answers.



*\begin{align}
\int_{0}^{1}{1 - x^{n} \over \ln\pars{x}}\,\dd x & =
-\int_{0}^{1}\pars{1 - x^{n}}\int_{0}^{\infty}x^{y}\,\dd y\,\dd x =
\int_{0}^{\infty}\int_{0}^{1}\pars{x^{n + y} - x^{y}}\,\dd x\,\dd y
\\[3mm] = &\
\int_{0}^{\infty}\pars{{1 \over y + n + 1} - {1 \over y + 1}}\,\dd y =
\fbox{$\ds{\ -\ln\pars{n + 1}\ }$}
\end{align}



*\begin{align}
\int_{0}^{1}{1 - x^{n} \over 1 - x} & =\int_{0}^{1}\sum_{k=1}^{n}x^{k - 1}
\,\dd x =
\sum_{k = 1}^{n}{1 \over k}\ \stackrel{\mathrm{def.}}{=}\
\fbox{$\ds{\ H_{n}\ }$}\quad n = 1,2,3,\ldots
\end{align}
Indeed, the integral defines/extends the Harmonic Number for
$\ds{~\Re\pars{n} > -1~}$ and, in general, by means of the Digamma Function
$\ds{\pars{~namely, H_{z} = \Psi\pars{z + 1} + \gamma~}}$.

A: Hint. One may set
$$
f(s):=\int_{0}^{1}x^s\left({1\over \ln{x}}+{1\over 1-x}\right)dx, \quad s>-1, \tag1
$$ then just differentiate getting
$$
f'(s)=\int_{0}^{1}x^s\left(1+{\ln x\over 1-x}\right)dx=\frac1{s+1}-\psi'\left(s+1\right), \quad s>-1, \tag2
$$where we have used the standard integral representation of the digamma function
$$
\int_{0}^{1}{1 - x^{s - 1} \over 1 - x}\,dx
\, = \psi (s)+ \gamma, \quad s>0.
$$
Integrating $(2)$, observing that $f(0)=\gamma$, one gets

$$
\int_{0}^{1}x^s\left({1\over \ln{x}}+{1\over 1-x}\right)dx=\log(s+1)-\psi\left(s+1\right), \quad s>-1, \tag3
$$ 

from which you deduce the value of your initial integral by recalling that 
$$
\psi(n+1)=H_n-\gamma.
$$
A: Your approach cannot work, because the series $J$ that you derived is incorrect:  the sum is not alternating, since $$\frac{x^n}{1-x} = \sum_{k=n}^\infty x^k,$$ which also reveals the more problematic issue that $J$ is not convergent and that the original integrand cannot be split in such a fashion.
Instead, write $$f_n(x) = x^n \left( \frac{1}{\log x} + \frac{1}{1-x} \right) = \frac{x^n-1}{\log x} + \frac{1}{\log x} + \frac{x^n}{1-x}.$$  Then consider $$g(n,x) = \int_{x=0}^1 \frac{x^n-1}{\log x} \, dx$$ and note $$\frac{dg}{dn} = \int_{x=0}^1 x^n \, dx = \frac{1}{1+n}$$ for (at least) $n \ge 0$.  Then $$g(n,x) = \int \frac{1}{1+n} \, dn = \log (1+n) + C,$$ and the additional condition that $g(0,x) = 0$ gives $$g(n,x) = \log(1+n).$$  This gives the desired middle term.  Now what remains is $$\int_{x=0}^1 f_n(x) \, dx  = \log (1+n) + \int_{x=0}^1 h_n(x) \, dx,$$ for which the harmonic part can be removed simply by noting $$\frac{x^n}{1-x} = \frac{1}{1-x} - \sum_{k=0}^{n-1} x^k,$$ hence $$\int_{x=0}^1 h_n(x) \, dx = \int_{x=0}^1 h_0(x) \, dx - \sum_{k=0}^{n-1} \frac{1}{k+1} = \int_{x=0}^1 h_0(x) \, dx - H_n.$$  Then all that remains is to prove $$\gamma = \int_{x=0}^1 h_0(x) \, dx = \int_{x=0}^1 \frac{1}{\log x} + \frac{1}{1-x} \, dx,$$ which is actually claimed in the Wikipedia article for the Euler-Mascheroni constant but I have left it as an exercise.
A: Let me propose a complete elementary answer based on $\zeta(s)\Gamma(s)$ :


*

*since $$\zeta(s) = \lim_{N \to \infty}\sum_{n=1}^N n^{-s}$$
it is clear that 


$$\lim_{s \to 1^+} \zeta(s) - \frac{1}{s-1} = \lim_{s \to 1^+} \lim_{N \to \infty}\sum_{n=1}^N n^{-s}- \int_1^N x^{-s}dx$$ $$ = \lim_{N \to \infty}\lim_{s \to 1^+}\sum_{n=1}^N n^{-s}- \int_1^N x^{-s}dx = \lim_{N \to \infty} H_N - \ln N = \gamma$$ 


*

*then use $n^{-s} \Gamma(s) = \int_0^\infty x^{s-1} e^{-nx} dx$ (change of variable $y = nx$) so that $\Gamma(s) \sum_{n=1}^N n^{-s}= \int_0^\infty x^{s-1} \sum_{n=1}^N e^{-nx} dx  = \int_0^\infty x^{s-1} \frac{1-e^{-(N+1)x}}{1-e^{-x}} dx$ and for $Re(s) > 1$ : 
$$\Gamma(s) \zeta(s) = \lim_{N \to \infty}   \int_0^\infty x^{s-1} \frac{1-e^{-(N+1)x}}{1-e^{-x}} dx = \int_0^\infty x^{s-1}\frac{e^{-x}}{1-e^{-x}}dx $$
and you get :
$$\gamma = \lim_{s \to 1^+} \Gamma(s) \left(\zeta(s) - \frac{1}{s-1}\right) = \lim_{s \to 1^+} \int_0^\infty x^{s-1}e^{-x}\left(\frac{1}{1-e^{-x}}-\frac{1}{x}\right)dx$$ $$ = \int_0^\infty e^{-x}\left(\frac{1}{1-e^{-x}}-\frac{1}{x}\right)dx $$


*

*finally : 


$$\int_0^1 x^n\left(\frac{1}{\ln x} + \frac{1}{1-x}\right)dx = \int_0^\infty e^{-(n+1)t}\left( \frac{1}{1-e^{-t}}-\frac{1}{t}\right)dt$$ $$= \lim_{s \to 1^+} \int_0^\infty t^{s-1}e^{-(n+1)t}\left( \frac{1}{1-e^{-t}}-\frac{1}{t}\right)dt$$ $$ = \lim_{s \to 1^+} \Gamma(s) \sum_{n=N+1}^\infty n^{-s} - (n+1)^{-s}\Gamma(s-1)$$ $$ = 
\lim_{s \to 1^+} \Gamma(s) (\zeta(s) - \sum_{n=1}^N n^{-s}) - (n+1)^{1-s}\Gamma(s-1)$$
and from $(n+1)^{1-s}\Gamma(s-1) = \Gamma(s)\frac{1+(1-s)\ln(n+1)+o(1-s)}{s-1} $ you get
$$\int_0^1 x^n\left(\frac{1}{\ln x} + \frac{1}{1-x}\right)dx = \lim_{s \to 1^+} \Gamma(s)\left(\zeta(s) - \sum_{n=1}^N n^{-s}-\frac{1+(1-s)\ln(n+1)+o(1-s)}{s-1}\right)$$ $$ = \color{red}{\gamma-H_n+\ln(n+1)}$$
A: If $n\ge 0$
$$\begin{align}
  & \int_{0}^{n}{{{x}^{t}}}dt=\int_{0}^{n}{{{e}^{t\ln x}}}dt=\left. \frac{{{e}^{t\ln x}}}{\ln x} \right|_{0}^{n}=\frac{{{e}^{n\ln x}}-1}{\ln x}=\frac{{{x}^{n}}-1}{\ln x} \\ 
\\
 & \int_{0}^{1}{\frac{{{x}^{n}}-1}{\ln x}}\,dx=\int_{0}^{1}{\int_{0}^{n}{{{x}^{t}}}dt}dx=\int_{0}^{n}{\int_{0}^{1}{{{x}^{t}}}dx}dt=\int_{0}^{n}{\frac{1}{t+1}}dt=\ln (n+1) \\ 
\end{align}$$
Indeed we proved

$$\color{red}{\int_{0}^{1}{\frac{{{x}^{n}}-1}{\ln x}}\,dx=\ln(n+1)}\tag 1$$

$$\int_{0}^{1}\frac{1-x^n}{1-x}dx=\int_{0}^{1}{(1+x+x^2+x^3+\cdots+x^{n-1})}dx=\sum\limits_{j=1}^{n}{\frac{1}{j}}=\color{red}{{H}_{n}}$$
thus

$$\color{red}{\int_{0}^{1}\frac{x^n-1}{1-x}dx=-{{H}_{n}}}\tag 2$$

on the other hand, we know 
$$\color{red}{\int_{0}^{1}\left(\frac{1}{\ln x}+\frac{1}{1-x}\right)dx=\gamma }\tag 3$$
 $(1) , (2)$ and $(3)$
$$I=\int_{0}^{1}x^n\left({1\over \ln{x}}+{1\over 1-x}\right)dx=\int_{0}^{1}\left({1\over \ln{x}}+{1\over 1-x}\right)dx+\int_{0}^{1}\frac{x^n-1}{\ln x} dx+\int_{0}^{1}\frac{x^n-1}{1-x}dx$$
therefore
$$\color{red}{I=\gamma+\ln(n+1)-H_n}$$
