Showing $\int_{0}^{1}{x^{2n}-x\over 1+x}\cdot{dx\over \ln{x}}=\ln\left({2\over \pi}\cdot{(2n)!!\over (2n-1)!!}\right)$ Integrate

$$I=\int_{0}^{1}{x^{2n}-x\over 1+x}\cdot{dx\over \ln{x}}=\ln\left({2\over \pi}\cdot{(2n)!!\over (2n-1)!!}\right)\tag1$$

$${x^{2n}-x\over 1+x}=\sum_{k=0}^{\infty}(-1)^k(x^{2n}-x)x^k\tag2$$
Sub $(2)$ into $(1)\rightarrow (3)$
$$I=\sum_{n=0}^{\infty}\int_{0}^{1}(x^{2n+k}-x^{k+1})\cdot{dx\over \ln{x}}\tag3$$
Frullani's theorem
$$\int_{0}^{1}(x^m-x^n)\cdot{dx\over \ln{x}}=\ln{m+1 \over n+1}\tag4$$
Apply $(4)$ to $(3)\rightarrow (5)$
$$I=\sum_{n=0}^{\infty}(-1)^n\ln\left({2n+k+1\over k+2}\right)\tag5$$
$${2\over \pi}={(2n-1)!!\over (2n)!}\prod_{k=0}^{\infty}\left({2n+k+1\over k+2}\right)^{(-1)^k}\tag6$$
 A: Enforce the substitution $x\to e^{-x}$ to write 
$$\begin{align}
I(n)&=\int_0^1 \frac{x^{2n}-x}{1+x}\frac{1}{\log(x)}\,dx\\\\
&=\int_0^\infty \frac{e^{-x}-e^{-2nx}}{x}\frac{e^{-x}}{1+e^{-x}}\,dx\\\\
&=\sum_{k=0}^\infty (-1)^k \int_0^\infty \frac{e^{-(k+2)x}-e^{-(k+2n+1)x}}{x}\\\\
&=\sum_{k=0}^\infty (-1)^k \log\left(\frac{k+2n+1}{k+2}\right)\\\\
&=\sum_{k=1}^\infty (-1)^{k-1} \log\left(\frac{k+2n}{k+1}\right)\\\\
\end{align}$$
Now, note that we can write the partial sum
$$\begin{align}
\sum_{k=1}^{2N} (-1)^{k-1} \log\left(\frac{k+2n}{k+1}\right)&=\sum_{k=1}^N  \log\left(\frac{2k-1+2n}{2k}\right)-\sum_{k=1}^N \log\left(\frac{2k+2n}{2k+1}\right)\\\\
&=\sum_{k=1}^N  \log\left(\frac{2k-1+2n}{2k+2n}\right)+\sum_{k=1}^N \log\left(\frac{2k+1}{2k}\right)\\\\
&=\sum_{k=n+1}^{n+N}  \log\left(\frac{2k-1}{2k}\right)+\sum_{k=1}^N \log\left(\frac{2k+1}{2k}\right)\\\\
&=\sum_{k=1}^{n+N}  \log\left(\frac{2k-1}{2k}\right)+\sum_{k=1}^{n+N} \log\left(\frac{2k+1}{2k}\right)\\\\
&-\sum_{k=1}^{n}  \log\left(\frac{2k-1}{2k}\right)-\sum_{k=N+1}^{n+N} \log\left(\frac{2k+1}{2k}\right)\\\\
&=\sum_{k=1}^{n+N}  \log\left(\frac{(2k-1)(2k+1)}{(2k)(2k)}\right)\\\\
&+\log\left(\frac{(2n)!!}{(2n-1)!!}\right)-\sum_{k=N+1}^{n+N} \log\left(\frac{2k+1}{2k}\right)\\\\
\end{align}$$
Recalling Wallis' Product, we see that 
$$\lim_{N\to \infty}\sum_{k=1}^{n+N}  \log\left(\frac{(2k-1)(2k+1)}{(2k)(2k)}\right)=-\log(\pi/2)$$
And since $\lim_{N\to \infty}\sum_{k=N+1}^{n+N} \log\left(\frac{2k+1}{2k}\right)=0$, we find
$$I(n)=\log\left(\frac{2}{\pi}\frac{(2n)!!}{(2n-1)!!}\right)$$
as was to be shown!!

APPENDIX:
In THIS ANSWER, I evaluated the integral 
$$J(n)=\int_0^1 \frac{x^{2n+1}-x}{1+x}\frac{1}{\log(x)}\,dx=\log\left(\frac{(2n+1)!!}{(2n)!!}\right)$$
by making use of the integral evaluated herein.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}
&\color{#f00}{\int_{0}^{1}{x^{2n} - x \over 1 + x}\,{\dd x\over \ln\pars{x}}} =
-\int_{0}^{1}{x^{2n} - x \over 1 + x}\
\overbrace{\int_{0}^{\infty}x^{y}\,\dd y}^{\ds{-\,{1 \over \ln\pars{x}}}}\
\,\dd x =
-\int_{0}^{\infty}\int_{0}^{1}{x^{2n + y} - x^{1 + y} \over 1 + x}\,\dd x\,\dd y
\\[3mm] = &\
\int_{0}^{\infty}\pars{\int_{0}^{1}{1 - x^{y + 2n} \over 1 + x}\,\dd x -
\int_{0}^{1}{1 - x^{y + 1} \over 1 + x}\,\dd x}\,\dd y\tag{1}
\end{align}

However, by using the well known digamma $\Psi$ function identity
$\ds{\left.\int_{0}^{1}{1 - t^{z - 1} \over 1 - t}\,\dd t
\,\right\vert_{\ \Re\pars{z}\ >\ 0} = \Psi\pars{z} + \gamma\quad}$
where $\gamma$ is the Euler-Mascheroni constant:
\begin{align}
\fbox{$\ds{\int_{0}^{1}{1 - x^{z} \over 1 + x}\,\dd x}$} &=
2\int_{0}^{1}{1 - x^{z} \over 1 - x^{2}}\,\dd x -
\int_{0}^{1}{1 - x^{z} \over 1 - x}\,\dd x
\\[3mm] & =
\int_{0}^{1}{x^{-1/2} - x^{z/2 - 1/2} \over 1 - x}\,\dd x -
\int_{0}^{1}{1 - x^{z} \over 1 - x}\,\dd x
\\[3mm] & =
\int_{0}^{1}{1 - x^{z/2 - 1/2} \over 1 - x}\,\dd x -
\int_{0}^{1}{1 - x^{-1/2} \over 1 - x}\,\dd x -
\int_{0}^{1}{1 - x^{z} \over 1 - x}\,\dd x
\\[3mm] & = \fbox{$\ds{%
\Psi\pars{{z \over 2} + \half} - \Psi\pars{\half} - \Psi\pars{z + 1} - \gamma}$}
\end{align}

we get, after replacing in $\pars{1}$,
\begin{align}
&\color{#f00}{\int_{0}^{1}{x^{2n} - x \over 1 + x}\,{\dd x\over \ln\pars{x}}}
\\[3mm] = &\
\int_{0}^{\infty}\bracks{\Psi\pars{{y \over 2} + n + \half} -
\Psi\pars{y + 2n + 1} - \Psi\pars{{y \over 2} + 1} + \Psi\pars{y + 2}}\,\dd y
\end{align}
Since
$\ds{\Psi\pars{z}\ \stackrel{\mbox{def.}}{=}\
     \totald{\ln\pars{\Gamma\pars{z}}}{z}}$:
\begin{align}
&\color{#f00}{\int_{0}^{1}{x^{2n} - x \over 1 + x}\,{\dd x\over \ln\pars{x}}} =
\left.\ln\pars{\Gamma^{2}\pars{y/2 + n + 1/2}\Gamma\pars{y + 2} \over \Gamma^{2}\pars{y/2 + 1}\Gamma\pars{y + 2n + 1}}\right\vert_{\ 0}^{\ \infty}
\\[3mm] = &\
\color{#f00}{%
\ln\pars{2^{1 - 2n}\,{\Gamma\pars{2n + 1} \over \Gamma^{2}\pars{n + 1/2}}}}
\end{align}
Could you simplify it ?.
