<s>[INST] 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.  
 [/INST] 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,
$\boxed{\displaystyle J=\frac{1}{2}\ln^2 2-\frac{1}{12}\pi^2}$
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