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.  

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.
