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\begin{align}
&\color{#f00}{%
\int_{0}^{1}{1 \over u + \lambda}\, \ln\pars{1 + u \over 1 - u}\,\dd u}
=
-\int_{1}^{2}{\ln\pars{u} \over \pars{1 - \lambda} - u}\,\dd u -
\int_{0}^{1}{\ln\pars{u} \over \pars{1 + \lambda} - u}\,\dd u
\end{align}
Then, integrals are of the form
\begin{align}
\int_{a}^{b}{\ln\pars{u} \over c - u}\,\dd u & =
\int_{a/c}^{b/c}{\ln\pars{cu} \over 1 - u}\,\dd u =
\left.\vphantom{\Large A}-\ln\pars{1 - u}\ln\pars{cu}\right\vert_{\ a/c}^{\ b/c}
+ \int_{a/c}^{b/c}\
\overbrace{{\ln\pars{1 - u} \over u}}^{\ds{-\,\mathrm{Li}_{2}'\pars{u}}}\
\,\dd u
\\[3mm] & = -\ln\pars{c - b \over c}\ln\pars{b} +
\ln\pars{c - a \over c}\ln\pars{a} -
\mathrm{Li}_{2}\pars{b \over c} + \mathrm{Li}_{2}\pars{a \over c}
\end{align}
$\ds{\large a = 1\,,\ b = 2\,,\ c = 1 - \lambda}$:
\begin{align}
\int_{1}^{2}{\ln\pars{u} \over \pars{1 - \lambda} - u}\,\dd u & =
-\ln\pars{\lambda + 1 \over \lambda - 1}\ln\pars{2} -
\mathrm{Li}_{2}\pars{2 \over 1 - \lambda} +
\mathrm{Li}_{2}\pars{1 \over 1 - \lambda}
\end{align}
$\ds{\large a = 0\,,\ b = 1\,,\ c = 1 + \lambda}$:
\begin{align}
\int_{0}^{1}{\ln\pars{u} \over \pars{1 + \lambda} - u}\,\dd u & =
-\mathrm{Li}_{2}\pars{1 \over 1 + \lambda}
\end{align}
\begin{align}
&\color{#f00}{%
\int_{0}^{1}{1 \over u + \lambda}\, \ln\pars{1 + u \over 1 - u}\,\dd u}
\\[3mm] = &\
\color{#f00}{\ln\pars{\lambda + 1 \over \lambda - 1}\ln\pars{2} +
\mathrm{Li}_{2}\pars{2 \over 1 - \lambda} -
\mathrm{Li}_{2}\pars{1 \over 1 - \lambda} +
\mathrm{Li}_{2}\pars{1 \over 1 + \lambda}}
\end{align}