Evaluation of the integral $\int_0^1 \frac{\ln(1 - x)}{1 + x}dx$ How can I evaluate the integral
$$\int_0^1 \frac{\ln(1 - x)}{1 + x}dx$$ 
I tried manipulating the known integral
$$\int_0^1 \frac{\ln(1 - x)}{x}dx = -\frac{\pi^2}{6}$$
but couldn't do anything with it.  
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$\ds{\int_{0}^{1}{\ln\pars{1 - x} \over 1 + x}\,\dd x = -\,{\pi^{2} \over 6}:\ {\large ?}}$

\begin{align}
&\color{#c00000}{\int_{0}^{1}{\ln\pars{1 - x} \over 1 + x}\,\dd x}
=\int_{0}^{1}{\ln\pars{x} \over 2 - x}\,\dd x
=\int_{0}^{1/2}{\ln\pars{2x} \over 1 - x}\,\dd x
\\[3mm]&=
\overbrace{\left.\vphantom{\Huge a}-\ln\pars{1 - x}\ln\pars{2x}\right\vert_{0}^{1/2}}
^{\ds{=\ 0}}\ +\
\int_{0}^{1/2}\ln\pars{1 - x}\,{1 \over x}\,\dd x
=\color{#c00000}{-\int_{0}^{1/2}{{\rm Li}_{1}\pars{x} \over x}\,\dd x}
\end{align}
  where $\ds{{\rm Li_{s}}\pars{z}}$ is the PolyLogarithm Function. We already used $\ds{{\rm Li_{1}}\pars{z} = -\ln\pars{1 - z}}$.

With the identity ( see the above mentioned link )
$\ds{{\rm Li_{s + 1}}\pars{z} = \int_{0}^{z}{{\rm Li_{s}}\pars{t} \over t}\,\dd t}$
we'll have:
$$
\color{#c00000}{\int_{0}^{1}{\ln\pars{1 - x} \over 1 + x}\,\dd x}
=\color{#c00000}{-{\rm Li_{2}}\pars{\half}}
$$
Also, ( see the above mentioned link )
$\ds{{\rm Li_{2}}\pars{\half} = {\pi^{2} \over 12} - \half\,\ln^{2}\pars{2}}$ which is a consequence of
Euler Reflection Formula
$\ds{{\rm Li_{2}}\pars{x} + {\rm Li_{2}}\pars{1 - x}
     ={\pi^{2} \over 6} -\ln\pars{x}\ln\pars{1 - x}}$.

$$
\color{#00f}{\large\int_{0}^{1}{\ln\pars{1 - x} \over 1 + x}\,\dd x
=\half\,\ln^{2}\pars{2} - {\pi^{2} \over 12}}
$$

A: You can use double integration:
$$\int\limits_0^1 {\frac{{\log \left( {1 - x} \right)}}{{1 + x}}dx}  = \int\limits_0^1 {\int\limits_0^{ - x} {\frac{{du \cdot dx}}{{\left( {1 + u} \right)\left( {1 + x} \right)}}} } $$
$$\int\limits_0^1 {\int\limits_0^x {\frac{{dm \cdot dx}}{{\left( {m - 1} \right)\left( {1 + x} \right)}}} } $$
Now make 
$$m = ux $$
$$\int\limits_0^1 {\int\limits_0^1 {\frac{{x \cdot du \cdot dx}}{{\left( {ux - 1} \right)\left( {1 + x} \right)}}} }  = \int\limits_0^1 {\int\limits_0^1 {\frac{{du \cdot dx}}{{\left( {ux - 1} \right)}}} }  - \int\limits_0^1 {\int\limits_0^1 {\frac{{du \cdot dx}}{{\left( {ux - 1} \right)\left( {1 + x} \right)}}} } $$
We have that (partial fraction decomposition)
$$\frac{1}{ \left( ux - 1 \right)\left( x + 1 \right) } = \frac{u}{ \left( u + 1 \right)\left( ux - 1 \right) } - \frac{1}{ \left( x + 1 \right)\left( u + 1 \right) }$$
So we get
$$\int\limits_0^1 {\int\limits_0^1 {\frac{{du \cdot dx}}{{\left( {ux - 1} \right)}}} }  - \int\limits_0^1 {\int\limits_0^1 {\frac{{u \cdot du \cdot dx}}{{\left( {ux - 1} \right)\left( {u + 1} \right)}}} }  + \int\limits_0^1 {\int\limits_0^1 {\frac{{du \cdot dx}}{{\left( {x + 1} \right)\left( {u + 1} \right)}}} } $$
Now:
$$\int\limits_0^1 {\int\limits_0^1 {\frac{{du \cdot dx}}{{\left( {ux - 1} \right)}}} }  = \int\limits_0^1 {\frac{{\log \left( {1 - u} \right)}}{u}} du =  - \frac{{{\pi ^2}}}{6}$$
$$\int\limits_0^1 {\int\limits_0^1 {\frac{{du\cdot dx}}{{\left( {x + 1} \right)\left( {u + 1} \right)}}} }  = {\log ^2}2$$
For our last one,note it is the integral we're looking for
$$\int\limits_0^1 {\int\limits_0^1 {\frac{{u\cdot du\cdot dx}}{{\left( {ux - 1} \right)\left( {u + 1} \right)}}} \mathop  = \limits^{ux = m} } \int\limits_0^1 {\int\limits_0^u {\frac{{dm\cdot du}}{{\left( {m - 1} \right)\left( {u + 1} \right)}}} } \mathop  = \limits^{m =  - x} \int\limits_0^1 {\int\limits_0^{ - u} {\frac{{dx\cdot du}}{{\left( {x + 1} \right)\left( {u + 1} \right)}}} }  = \int\limits_0^1 {\frac{{\log \left( {1 - u} \right)}}{{ {u + 1} }}} du$$
We get
$$\int\limits_0^1 {\frac{{\log \left( {1 - u} \right)}}{{ {u + 1} }}} du = {\log ^2}2 - \frac{{{\pi ^2}}}{6} - \int\limits_0^1 {\frac{{\log \left( {1 - u} \right)}}{{ {u + 1} }}} du$$
or
$$\int\limits_0^1 {\frac{{\log \left( {1 - u} \right)}}{{{u + 1} }}} du = \frac{{{{\log }^2}2}}{2} - \frac{{{\pi ^2}}}{{12}}$$
as desired.
A: Maple says it's $${(\log2)^2\over2}-{\pi^2\over12}$$ To get there, I think you will have to understand how the known integral you cite was established, and then use the same ideas to do yours (perhaps after first following Emile's calculations). 
A: Following is an elementary proof.
I assume only that $\displaystyle \int_0^1 \frac{\ln x}{1-x}dx=-\frac{\pi^2}{6}$
\begin{align}J&=\int_0^1 \frac{\ln(1-x)}{1+x}dx\\
&\overset{y=\frac{1-x}{1+x}}=\int_0^1 \frac{\ln\left(\frac{2y}{1+y}\right)}{1+y}dy\\
&=\int_0^1 \frac{\ln\left(\frac{2}{1+y}\right)}{1+y}dy+\int_0^1 \frac{\ln t}{1+t}dt\\
&\overset{u=\frac{1-y}{1+y}}=\int_0^1 \frac{\ln\left(1+u\right)}{1+u}du+\int_0^1 \frac{\ln t}{1+t}dt\\
&=\frac{1}{2}\ln^2 2+\int_0^1 \frac{\ln t}{1+t}dt\\
\int_0^1 \frac{\ln t}{1+t}dt&=\int_0^1 \frac{\ln x}{1-x}dx-\int_0^1 \frac{2t\ln t}{1-t^2}dt\\
&\overset{w=t^2}=\int_0^1 \frac{\ln x}{1-x}dx-\frac{1}{2}\int_0^1 \frac{\ln w}{1-w}dw\\
&=\frac{1}{2}\int_0^1 \frac{\ln x}{1-x}dx\\
&=-\frac{1}{12}\pi^2
\end{align}
Therefore,
$\boxed{\displaystyle J=\frac{1}{2}\ln^2 2-\frac{1}{12}\pi^2}$
A: You can use the integral you want to use, and the Dilogarithm function as mentioned in the comments.
Below we give a complete proof, including a derivation of the value of the integral you wanted to use.
The Dilogarithm function is defined as
$$\text{Li}_2(z) = -\int_{0}^{z} \frac{\log (1-x)}{x} \text{dx} = \sum_{n=1}^{\infty} \frac{z^n}{n^2}, \quad |z| \le 1$$ 
The integral which you want to use is $\displaystyle -\text{Li}_2(1)$.
Note that $\displaystyle \text{Li}_2(1) = \sum_{n=1}^{\infty} \frac{1}{n^2} = \zeta(2) = \frac{\pi^2}{6}$. (For multiple proofs of that, see here: Different methods to compute $\sum\limits_{k=1}^\infty \frac{1}{k^2}$)
In your integral(whose value you want), make the substitution $\displaystyle x = 2t -1$ and we get
$$\int_{\frac{1}{2}}^{1} \frac{\log (2(1-t))}{t} \text{dt} = \log^2 2 + \int_{\frac{1}{2}}^{1} \frac{\log (1-t)}{t} \text{dt} = \log^2 2 + \text{Li}_2 \left(\frac{1}{2} \right) - \text{Li}_2(1) $$
Now the Dilogarithm function also satisfies the identity
$$\text{Li}_2(x) + \text{Li}_2(1-x)  = \frac{\pi^2}{6}-\log x \log (1-x), 0 \lt x \lt 1$$
This identity can easily be proven by just differentiating and using the value of $\displaystyle \text{Li}_2(1)$: 
$$\text{Li}_2'(x) - \text{Li}_2'(1-x) = -\frac{\log (1-x)}{x} + \frac{\log x}{1-x} = (-\log x \log (1-x))'$$
and so 
$$\text{Li}_2(x) + \text{Li}_2(1-x)  = C -\log x \log (1-x), 0 \lt x \lt 1$$
Taking limits as $\displaystyle x \to 1$ gives us $\displaystyle C = \frac{\pi^2}{6}$.
Thus
$$\text{Li}_2(x) + \text{Li}_2(1-x)  = \frac{\pi^2}{6}-\log x \log (1-x), 0 \lt x \lt 1$$
Setting $\displaystyle x = \frac{1}{2}$ gives us the value of $\displaystyle \text{Li}_2\left(\frac{1}{2}\right) = \frac{\pi^2}{12} - \frac{\log^2 2}{2}$
Thus your integral is
$$\log^2 2 + \text{Li}_2 \left(\frac{1}{2} \right) - \text{Li}_2(1) = \frac{\log^2 2}{2} - \frac{\pi^2}{12}$$
A: Note: this is not a complete solution, but may serve as a starter
First let $2u=x+1$ and thus $2du=dx$. Then we get:
$$\int_0^1\frac{\ln(1-x)}{1+x}dx=\int_{\frac{1}{2}}^1\frac{\ln(2-2u)}{2u}2du$$
$$=\int_{\frac{1}{2}}^1\frac{\ln(2(1-u))}{u}du=\int_{\frac{1}{2}}^1\frac{\ln2+\ln(1-u)}{u}du$$
$$=\int_{\frac{1}{2}}^1\frac{\ln2}{u}du+\int_{\frac{1}{2}}^1\frac{\ln(1-u)}{u}du$$
A: With subbing $1-x=y$ we have
$$\int_0^1\frac{\ln^a(1-x)}{1+x}dx=\int_0^1\frac{\ln^a(y)}{2-y}dy$$
$$=\sum_{n=1}^\infty\frac{1}{2^n}\int_0^1 y^{n-1}\ln^a(y)dy$$
$$=(-1)^aa!\sum_{n=1}^\infty\frac{1}{2^nn^{a+1}}=(-1)^aa!\text{Li}_{a+1}\left(\frac12\right)$$
Some cases:
By using $\text{Li}_2\left(\frac12\right)=\frac12\zeta(2)-\frac12\ln^2(2)$ and $\text{Li}_3\left(\frac12\right)=\frac78\zeta(3)-\frac12\ln(2)\zeta(2)+\frac16\ln^3(2)$  we have
\begin{align}
\int_0^1\frac{\ln(1-x)}{1+x}\ dx=-\text{Li}_{2}\left(\frac12\right)=-\frac12\zeta(2)+\frac12\ln^2(2)\label{ln(1-x)/(1+x)}
\end{align}
\begin{align}
\int_0^1\frac{\ln^2(1-x)}{1+x}\ dx=2\text{Li}_{3}\left(\frac12\right)=\frac74\zeta(3)-\ln(2)\zeta(2)+\frac13\ln^3(2)\label{ln^2(1-x)/(1+x)}
\end{align}
\begin{align}
\int_0^1\frac{\ln^3(1-x)}{1+x}\ dx=-6\text{Li}_{4}\left(\frac12\right)\label{ln^3(1-x)/(1+x)}
\end{align}
\begin{align}
\int_0^1\frac{\ln^4(1-x)}{1+x}\ dx=24\text{Li}_{5}\left(\frac12\right)\label{ln^4(1-x)/(1+x)}
\end{align}
