Closed form $\int_{a}^b \cfrac{1}{(1+x) \, \left[\ln(1-x)-\ln(1+x)\right]} \, \mathrm{d}x$ I am encountering an integral which involves logarithms, in particular,
\begin{equation}
\int_{a}^b \cfrac{1}{(1+x) \, \left[\ln(1-x)-\ln(1+x)\right]} \, \mathrm{d}x,
\end{equation}
where $a$ and $b$ are finite real numbers.
Does this integral have an closed form solution ? It seems that integration by parts does not work...
 A: As it has already been mentioned in the comment section, the simple substitution $t=\ln\dfrac{1+x}{1-x}$ 
leads to an integral of the form $\displaystyle\int\frac{dt}{t~(e^t+1)},~$ which does not possess a closed form expression 
in terms of elementary functions and constants: see Liouville's theorem and the Risch algorithm for more details. However, its definite counterpart, when evaluated over the entire positive real 
semiaxis, yields $\displaystyle\int_0^\infty\frac{t^n}{e^t+1}~=~n!~\eta(n+1)~$ for $n>-1,$ and $~\displaystyle\int_0^\infty\frac{t^n}{e^t-1}=n!~\zeta(n+1)~$ for 
$n>0:~$ see the Riemann $\zeta$ and Dirichlet $\eta$ functions for more information. Unfortunately, in our case $n=-1$, which means that the integral diverges, due to the fact that the $\Gamma$ function has a pole at the origin.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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Hint

With $\ds{x = {1 - t \over 1 + t}}$:

\begin{align}
\int_{a}^{b}{1 \over \pars{1 + x}\bracks{\ln\pars{1 - x} - \ln\pars{1 + x}}}
\,\dd x =
-\int_{\pars{1- a}/\pars{1 + a}}^{\pars{1 - b}/\pars{1 + b}}
{\dd t \over \pars{1 + t}\ln\pars{t}}
\end{align}

A further procedure requires the particular values of $\ds{a}$ and $\ds{b}$. A useful expression is
  $\ds{\left.\int_{0}^{\infty}p^{q}\,\dd q\,\right\vert_{\ p\ \in\ \pars{0,1}} =
-\,{1 \over \ln\pars{p}}}$. 

