Evaluating $\int_{0}^{1}\frac{1-x}{1+x}\frac{\mathrm dx}{\ln x}$ Some time ago I came across  to the following integral:  
$$I=\int_{0}^{1}\frac{1-x}{1+x}\frac{\mathrm dx}{\ln x}$$ What are the hints on how to compute this integral?
 A: I tried with "Differentiation under integration sign":
$$J(\alpha)=\int_0^1\frac{1-x^\alpha}{1+x}\frac{dx}{\ln x}\quad \alpha>0$$
Then, as usual, $$\frac{dJ}{d\alpha}=-\int_0^1\frac{x^\alpha}{1+x}dx=-f(\alpha),\text{(say)}$$
Then, integrating $$J(\alpha)=-\int f(\alpha)d\alpha+c$$
I used Mathematica to evaluate $f(\alpha)$ (it gives difference of two Harmonic numbers) and its integral. The integral is just $\ln\frac{\Gamma(\frac{2m+1}{2})}{\Gamma(\frac{m+1}{2})}$. Note that $J(0)=0$ and putting $\alpha=1$, the result follows.
A: Consider
$$
\mathcal{I}(\alpha)=\int_0^1\frac{1-x^\alpha}{(1+x)\ln x}\ dx.\tag1
$$
Differentiating $(1)$ with respect $\alpha$ yields
\begin{align}
\frac{d\mathcal{I}}{d\alpha}&=\int_0^1\frac{\partial}{\partial\alpha}\left[\frac{1-x^\alpha}{(1+x)\ln x}\right]\ dx\\
&=-\int_0^1\frac{x^\alpha}{1+x}\ dx\\
&=-\int_0^1\sum_{k=0}^\infty(-1)^kx^{\alpha+k}\ dx\\
&=-\sum_{k=0}^\infty(-1)^k\int_0^1x^{\alpha+k}\ dx\\
&=-\sum_{k=0}^\infty\frac{(-1)^k}{\alpha+k+1}.\tag2
\end{align}
Now consider polygamma function
$$
\psi_n(z)=\frac{d^{n+1}}{dz^{n+1}}\ln\Gamma(z)\tag3
$$
and
$$
\psi_n(z)=(-1)^{n+1}n!\sum_{k=0}^\infty\frac{1}{(z+k)^{n+1}}.\tag4
$$
Hence by using $(4)$ we obtain
$$
\sum_{k=0}^\infty\frac{(-1)^{k}}{(z+k)^{n+1}}=\frac1{(-2)^{n+1}n!}\left[\psi_n\left(\frac{z}{2}\right)-\psi_n\left(\frac{z+1}{2}\right)\right].\tag5
$$
Using $(3)$ and $(5)$ then $(2)$ becomes
\begin{align}
\frac{d\mathcal{I}}{d\alpha}&=\frac12\left[\psi_0\left(\frac{\alpha+1}{2}\right)-\psi_0\left(\frac{\alpha+2}{2}\right)\right]\\
\mathcal{I}(\alpha)&=\frac12\int\left[\psi_0\left(\frac{\alpha+1}{2}\right)-\psi_0\left(\frac{\alpha+2}{2}\right)\right]\ d\alpha\\
&=\ln\Gamma\left(\frac{\alpha+1}{2}\right)-\ln\Gamma\left(\frac{\alpha+2}{2}\right)+C\\
&=\ln\Gamma\left(\frac{\alpha+1}{2}\right)-\ln\Gamma\left(\frac{\alpha+2}{2}\right)-\frac12\ln\pi,\tag6
\end{align}
where $\mathcal{I}(0)=0$ and $C=-\ln\Gamma\left(\frac{1}{2}\right)$.

Thus
\begin{align}
\int_0^1\frac{1-x}{(1+x)\ln x}\ dx&=\mathcal{I}(1)\\
&=\ln\Gamma\left(1\right)-\ln\Gamma\left(\frac{3}{2}\right)-\frac12\ln\pi\\
&=\color{blue}{\ln\left(\frac{2}{\pi}\right)}.
\end{align}
A: $\newcommand{\+}{^{\dagger}}
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\begin{align}
I & \equiv \bbox[#ffd, 5px]{\int_{0}^{1}{1 - x \over 1 + x}
\,{\dd x \over \ln\pars{x}}} =
\int_{0}^{1}{1 \over 1 + x}\ \overbrace{\bracks{-\int_{0}^{1}x^{t}\,\dd t}}
^{\ds{1 - x \over \ln\pars{x}}}\ \,\dd x =
-\int_{0}^{1}\int_{0}^{1}{x^{t} \over 1 + x}\,\dd t\,\dd x
\\[5mm] & =
-\int_{0}^{1}\int_{0}^{1}{x^{t} - x^{t + 1} \over 1 - x^{2}}\,\dd x\,\dd t
\,\,\,\stackrel{x^{2}\ \mapsto\ x}{=}\,\,\,
-\,{1 \over 2}\int_{0}^{1}\int_{0}^{1}{x^{t/2 - 1/2}\,\,\, -\, x^{t/2} \over
1 - x}\,\dd x\,\dd t
\\[5mm] & =
{1 \over 2}\int_{0}^{1}\bracks{%
\int_{0}^{1}{1 - x^{t/2 - 1/2} \over 1 - x}\,\dd x -
\int_{0}^{1}{1 - x^{t/2} \over 1 - x}\,\dd x}\,\dd t
\\[5mm] & =
{1 \over 2}\int_{0}^{1}\bracks{%
\Psi\pars{{t \over 2} + {1 \over 2}} -
\Psi\pars{{t \over 2} + 1}}\,\dd t\label{1}\tag{1}
\\[5mm] & =
\left.\ln\pars{\Gamma\pars{t/2 + 1/2} \over \Gamma\pars{t/2 + 1}}
\right\vert_{0}^{1} =
\ln\pars{{\Gamma\pars{1} \over \Gamma\pars{3/2}}
\,{\Gamma\pars{1} \over \Gamma\pars{1/2}}} =
\ln\pars{{1 \over \root{\pi}/2}\,{1 \over \root{\pi}}}
\\[5mm] & =
\bbox[10px, border:1px groove navy]{\ln\pars{2 \over \pi}}
\approx -0.4514 \\ &
\end{align}
(\ref{1}): See $\ds{\color{black}{\bf 6.3.22}}$ in A & S Table.
A: You may like this method. Note
$$ \int_0^1x^ada=-\frac{1-x}{\ln x}, \text{ for }x>0, H_a=\sum_{k=1}^\infty\frac{a}{k(a+k)}, $$
and hence
\begin{eqnarray}
\int_0^1\frac{1-x}{1+x}\frac{dx}{\ln x}&=&-\int_0^1\int_0^1\frac{x^a}{1+x}dadx\\
&=&-\int_0^1\int_0^1\frac{x^a}{1+x}dxda\\
&=&-\frac{1}{2}\int_0^1(H_{\frac{a}{2}}-H_{\frac{a-1}{2}})da\\
&=&-\frac{1}{2}\int_0^1\sum_{k=1}^\infty\left(\frac{a}{k(a+2k)}-\frac{a-1}{k(a+2k-1)}\right)da\\
&=&-\frac{1}{2}\sum_{k=1}^\infty\left(4\ln k-2\ln(k^2-\frac{1}{4})\right)\\
&=&-\frac{1}{2}\ln\prod_{k=1}^\infty\frac{k^4}{(k^2-\frac{1}{4})^2}\\
&=&-\prod_{k=1}^\infty\frac{k^2}{k^2-\frac{1}{4}}\\
&=&-\ln\left(\frac{\pi}{2}\right).
\end{eqnarray}
In the last step, we used the Wallis Formula from here.
A: Make the substitution $x=e^{-y}$ and do a little algebra to get the value of the integral to be
$$\int_0^{\infty} \frac{dy}{y} \frac{e^{-y} - e^{-2 y}}{1+e^{-y}} $$
Now Taylor expand the denominator and get
$$\int_0^{\infty} \frac{dy}{y} (e^{-y} - e^{-2 y}) \sum_{k=0}^{\infty} (-1)^k e^{-k y} $$
If we can reverse the order of sum and integral, we get
$$ \sum_{k=0}^{\infty} (-1)^k \int_0^{\infty} \frac{dy}{y} (e^{-(k+1) y} - e^{-(k+2) y}) $$
The integral inside may be evaluated exactly, and the result is the sum
$$ \sum_{k=0}^{\infty} (-1)^k \log{\frac{k+1}{k+2}} $$
$$ = \lim_{n \rightarrow \infty} \; \log{\frac{\frac{1}{2} \frac{3}{4} \ldots \frac{2 n-1}{2 n}}{\frac{2}{3} \frac{5}{6} \ldots \frac{2 n+1}{2 n+2}}} $$
$$ = \log{\left ( \frac{2}{\pi} \right )} $$
