Closed form for $\int_0^1 \frac{1}{u + \lambda} \ln \left(\frac{1 + u}{1 - u} \right) ~d u$ The parameter $\lambda$ is complex and  it's not on the real axis.
There are some similar cases:
Help me evaluate $\int_0^1 \frac{\log(x+1)}{1+x^2} dx$
Evaluate $\int_0^1 \frac{\ln(1+bx)}{1+x} dx $
Evaluation of the integral $\int_0^1 \frac{\ln(1 - x)}{1 + x}dx$

Supplement: How to figure out the two integal $I_1$ and $I_2$
This integral should be separated into two parts for the convergence condition of the complex parameter $\lambda$, let us consider the two integrals with a complex parameter :
$\mathrm{Im}\{\lambda\} \ne 0$ , and its real part is limited by integral: 
$$
I_1 = \int_0^1 \frac{\ln(1 - u)}{\lambda + u}\, d u 
$$
Mathematica gave the result  $I_1 = \mathrm{Li}_2 \big(\frac{\lambda}{1 + \lambda}\big)- \ln \left( \frac{\lambda}{1+\lambda} \right)\, \ln(1 + \lambda) - \frac{\pi^2}{6}$
$$
I_2 = \int_0^1 \frac{\ln(1 + u)}{\lambda + u}\, du = - \mathrm{Li}_2\big(\frac{1}{1 - \lambda}\big) + \mathrm{Li}_2 \big(\frac{2}{1 - \lambda}\big) + \ln 2 \, \ln \left(\frac{\lambda +1 }{\lambda - 1} \right)
$$
 A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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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}
A: \begin{align}
\int_0^1 \frac{1}{u + \lambda} \ln \frac{1 + u}{1 - u} du
\overset{u=\frac{1-t}{1+t} }=& -\int_0^1 \frac{\ln t}{[(1+ \lambda)-(1-\lambda)t](1+t)}dt\\
=& -\int_0^1 \bigg(\frac{\frac{1-\lambda}{1+\lambda}\ln t}{1-\frac{1-\lambda}{1+ \lambda}t}
+\frac{\ln t}{1+t}\bigg)\ dt\\
=&\ \text{Li}_2(\frac{1-\lambda}{1+\lambda} ) +\frac{\pi^2}{12}\\
\end{align}
A: Take the second integral as a demonstration. 
\begin{aligned}
I_2 &= \int_0^1 \frac{\ln (1 + u)}{\lambda + u} \\
 & \Downarrow \quad x:= 1 + u \\
&= \int_1^2 \frac{\ln x}{\underbrace{\lambda - 1}_{z} + x} d x= \int_1^2 \frac{\ln x}{z +x} dx \\
 &= \ln x \, \ln\big(1 + \frac{x}{z} \big) \bigg|_{x=1}^{x=2}  - \int_1^2 \frac{d x}{x} \ln \big(1 + \frac{x}{z} \big) \\
&= \ln 2\, \ln \left(\frac{\lambda + 1}{\lambda -1} \right) - \int_0^1 dt \,\int_1^2  \frac{dx}{z + tx}
\end{aligned}
Focus the last integral
\begin{aligned}
- \int_0^1 dt \,\int_1^2  \frac{dx}{z + tx} &= - \int_0^1 \frac{dt}{t} \,\int_1^2  \frac{t \, dx}{z + tx} \\
&= -\int_0^1 \frac{dt}{t} \, \ln \big(z + tx \big) \bigg|_{x=1}^{x=2} \\
 &= - \int_0^1 \frac{dt}{t} \, \ln\big[ 1 - \big(-\frac{2}{z}\big) \big] 
+ \int_0^1 \frac{dt}{t} \, \ln \big[ 1 - \big( - \frac{1}{z}\big) \big] \\
 &= \mathrm{Li}_2 \big(\frac{2}{1 - \lambda }\big) - \mathrm{Li}_2\big(\frac{1}{1 - \lambda} \big)
\end{aligned}
Hence: $\displaystyle I_2 = \mathrm{Li}_2 \big(\frac{2}{1 - \lambda }\big) - \mathrm{Li}_2\big(\frac{1}{1 - \lambda} \big) + \ln2\, \ln\left(\frac{\lambda+1}{\lambda-1}\right)$
Similarly, $I_1$ can be figured out.
