Some time ago I came across to the following integral: I=∫011−x1+xdxln⁡x What are the hints...

Make the substitution ${\displaystyle x=e^{-y}}$ and do a little algebra to get the value of the integral to be
${\displaystyle {\int }_0^{\mathrm{\infty }}\frac{dy}{y}\frac{e^{-y}-e^{-2y}}{1+e^{-y}}}$
Now Taylor expand the denominator and get
${\displaystyle {\int }_0^{\mathrm{\infty }}\frac{dy}{y}(e^{-y}-e^{-2y})\sum _{k=0}^{\mathrm{\infty }}{(-1)}^k{e}^{-ky}}$
If we can reverse the order of sum and integral, we get
${\displaystyle \sum _{k=0}^{\mathrm{\infty }}{(-1)}^k{\int }_0^{\mathrm{\infty }}\frac{dy}{y}(e^{-(k+1)y}-e^{-(k+2)y})}$
The integral inside may be evaluated exactly, and the result is the sum
${\displaystyle \sum _{k=0}^{\mathrm{\infty }}{(-1)}^k\log \frac{k+1}{k+2}}$
${\displaystyle =\underset{n\to \mathrm{\infty }}{lim}\log \frac{\frac{12}{\frac{34}{\dots }}\frac{2n-1}{2n}}{\frac{23}{\frac{56}{\dots }}\frac{2n+1}{2n+2}}}$
${\displaystyle =\log \left(\frac{2}{\pi }\right)}$

I tried with Differentiation under integration sign:
${\displaystyle J\left(\alpha \right)={\int }_0^1\frac{1-x^{\alpha }}{1+x}\frac{dx}{\ln x}\alpha >0}$
Then as usual,
${\displaystyle \frac{dJ}{d\alpha }=-{\int }_0^1\frac{x^{\alpha }}{1+x}dx=-f\left(\alpha \right)}$
Then, integrating
${\displaystyle J\left(\alpha \right)=-\int f\left(\alpha \right)d\alpha +C}$
I used Mathematica to evaluate ${\displaystyle f\left(\alpha \right)}$ (it gives difference of two Harmonic numbers) and its integral. The integral is just ${\displaystyle \ln \frac{\mathrm{\Gamma }\left(\frac{2m+1}{2}\right)}{\mathrm{\Gamma }\left(\frac{m+1}{2}\right)}}$. Note that ${\displaystyle J\left(0\right)=0}$ and putting ${\displaystyle \alpha =1}$, the result follows.

Consider
$I(\alpha )={\int }_0^1\frac{1-x^{\alpha }}{(1+x)\ln x}dx$ (1)
Differentiating (1) with respect $\alpha$ yields
$$\begin{gathered} \frac{dI}{d\alpha }={\int }_0^1\frac{d}{d\alpha }[\frac{1-x^{\alpha }}{(1+x)\ln x}]dx \\ =-{\int }_0^1\frac{x^{\alpha }}{1+x}dx \\ =-{\int }_0^1\sum _{k=0}^{\mathrm{\infty }}(-1{)}^k{x}^{\alpha +k}dx \\ =-\sum _{k=0}^{\mathrm{\infty }}(-1{)}^k{\int }_0^1{x}^{\alpha +k}dx \\ =-\sum _{k=0}^{\mathrm{\infty }}\frac{(-1{)}^k}{\alpha +k+1} \end{gathered}$$ (2)
Now consider polygamma function
${\psi }_n(z)=\frac{d^{n+1}}{d{z}^{n+1}}\ln \mathrm{\Gamma }(z)$ (3)
and
${\psi }_n(z)=(-1{)}^{n+1}n!\sum _{k=0}^{\mathrm{\infty }}\frac{1}{(z+k{)}^{n+1}}$ (4)
Hence by using (4) we obtain
$\sum _{k=0}^{\mathrm{\infty }}\frac{(-1{)}^k}{(z+k{)}^{n+1}}=\frac{1}{(-2{)}^{n+1}n!}[{\psi }_n(\frac{z}{2})-{\psi }_n(\frac{z+1}{2})]$ (5)
Using (3) and (5) then (2) becomes
$$\begin{gathered} \frac{dI}{d\alpha }=\frac{1}{2}[{\psi }_0(\frac{\alpha +1}{2})-{\psi }_0(\frac{\alpha +2}{2})] \\ I(\alpha )=\frac{1}{2}\int [{\psi }_0(\frac{\alpha +1}{2})-{\psi }_0(\frac{\alpha +2}{2})]d\alpha \\ =\ln \mathrm{\Gamma }(\frac{\alpha +1}{2})-\ln \mathrm{\Gamma }(\frac{\alpha +2}{2})+C \\ =\ln \mathrm{\Gamma }(\frac{\alpha +1}{2})-\ln \mathrm{\Gamma }(\frac{\alpha +2}{2})+\frac{1}{2}\ln \pi \\ \text{where}I(0)=0\text{and}C=-\ln \mathrm{\Gamma }(\frac{1}{2}) \\ \text{Thus} \\ {\int }_0^1\frac{1-x}{(1+x)\ln x}dx=I(1) \\ =\ln \mathrm{\Gamma }(1)-\ln \mathrm{\Gamma }(\frac{3}{2})-\frac{1}{2}\ln \pi \\ =\ln (\frac{2}{\pi }) \end{gathered}$$