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Calculate: $$\int_{0}^{1}\frac{\log x}{2 - x} dx$$

I've done a lot of research here in the community. I tried varying variants, but I did not get anything. The problem is the $2$ present in the integral.

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Substituting $\dfrac{x}{2-x}=t\Rightarrow x=\dfrac{2t}{1+t}$ we get: $$I=\int_0^1 \frac{x\ln x}{{2-x}}\frac{dx}{x}=\int_0^1 \ln \left(\frac{2t}{1+t}\right) \frac{dt}{1+t} $$ $$=\int_0^1 \frac{\ln2 +\ln t -\ln(1+t)}{1+t}dt$$ $$=\ln2 \ln(1+t)\bigg|_0^1 +\int_0^1 \frac{\ln t}{1+t}dt - \frac12 \ln^2 (1+t)\bigg|_0^1 $$ $$=\ln^2 2 -\frac{\pi^2}{12} -\frac12 \ln^22=\frac{\ln^2 2}{2}-\frac{\pi^2}{12}$$ The second integral can be found here for example.


One can also make use of the Dilogarithm values.$$I=\int_0^1 \frac{\ln x}{2-x}dx=\frac12 \int_0^1 \frac{\ln x}{1-\frac{x}{2}}dx=\frac12 \sum_{n= 0}^\infty \frac{1}{2^n}\int_0^1 x^n \ln xdx$$$$=-\frac12 \sum_{n= 0}^\infty \frac{1}{2^n} \frac{1}{(n+1)^2}=-\sum_{n= 1}^\infty \frac{1}{2^n} \frac{1}{n^2}=-\operatorname{Li}_2 \left(\frac12 \right) =\frac{\ln^2 2}{2}-\frac{\pi^2}{12}$$

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    $\begingroup$ Good Job. Thank's. $\endgroup$ – Mathsource Feb 15 at 1:25
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\begin{align}J&=\int_{0}^{1}\frac{\log x}{2 - x} dx\\ \end{align}

Perform the change of variable $x=2y$,

\begin{align}J&=\int_{0}^{\frac{1}{2}}\frac{\log(2y)}{1 - y} dy\\ &=\ln 2\left[-\ln(1-x)\right]_{0}^{\frac{1}{2}}+\int_{0}^{\frac{1}{2}}\frac{\log y}{1 - y} dy\\ &=\ln ^2 2+\int_{0}^{\frac{1}{2}}\frac{\log y}{1 - y} dy\\ &=\ln ^2 2+\left[-\ln(1-y)\ln y\right]_{0}^{\frac{1}{2}}+\int_{0}^{\frac{1}{2}}\frac{\ln(1-y)}{y}\,dy\\ &=\int_{0}^{\frac{1}{2}}\frac{\ln(1-y)}{y}\,dy\\ \end{align}

Perform the change of variable $u=1-y$,

\begin{align}J&= \int_{\frac{1}{2}}^1\frac{\ln u}{1-u}\,du\\ &=\int_{0}^1\frac{\ln u}{1-u}\,du-\int_0^{\frac{1}{2}}\frac{\ln u}{1-u}\,du\\ &=\int_{0}^1\frac{\ln u}{1-u}\,du+\ln^2 2-J\\ \end{align}

Therefore,

\begin{align}J&=\frac{1}{2}\int_{0}^1\frac{\ln u}{1-u}\,du+\frac{1}{2}\ln^2 2\\ &=\frac{1}{2}\int_0^1\left( \sum_{n=0}^\infty u^n\right)\ln u\,du+\frac{1}{2}\ln^2 2\\ &=\frac{1}{2}\sum_{n=0}^\infty\left(\int_0^1 u^n\ln u\,du\right)+\frac{1}{2}\ln^2 2\\ &=\frac{1}{2}\ln^2 2-\frac{1}{2}\sum_{n=0}^\infty\frac{1}{(n+1)^2}\\ &=\boxed{\frac{1}{2}\ln^2 2-\frac{1}{2}\zeta(2)}\\ \end{align}

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$$I=\frac{1}{2}\int^{1}_{0}\frac{\ln x}{1-\frac{x}{2}}dx=\frac{1}{2}\int^{1}_{0}\ln (x)\sum^{\infty}_{n=0}\bigg(\frac{x}{2}\bigg)^n$$

$$I=\frac{1}{2}\sum^{\infty}_{n=0}\frac{1}{2^n}\int^{1}_{0}x^n \ln(x)dx$$

$$I=-\frac{1}{2}\sum^{\infty}_{n=0}\frac{1}{2^n}\frac{1}{(n+1)^2}=-\sum^{\infty}_{n=0}\frac{1}{2^n}\frac{1}{(n+1)^2}$$

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    $\begingroup$ (x/2)^n, not (x/2) $\endgroup$ – Seth Feb 15 at 1:00

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