Solving: $\int_0^1\frac{\ln(1-x)}{(1+x)}dx$
$$\begin{aligned} & K=\int_0^1 \frac{\ln (1-x)}{1+x} d x-\int_0^1 \frac{\ln (x)}{1+x} d x \\ & I=\int_0^1 \frac{\ln (1-x)}{(1+x)} d x =\int_1^2 \frac{\ln \left(1-\frac{t}{2}\right)+\ln (2)}{t} d t \\ &=\int_{\frac{1}{2}}^1 \frac{\ln (1-m)}{m} d t+\int_1^2 \frac{\ln (2)}{t} d t \\ &=-\left[Li_2(1)-L i_2\left(\frac{1}{2}\right)\right]+\ln^{2}(2) \\ &J=\int_0^2 \frac{\ln (x)}{1+x} d x=\left.\ln (x) \ln (1+x)\right|_0 ^1+Li_2(-1) =\left(\frac{-\pi^2}{12}\right) \\ & K=I-J\\&K=\frac{\pi^2}{12}-\left[\frac{\pi^2}{6}-\left(\frac{\pi^2}{12}-\frac{\ln ^2(2)}{2}\right)\right]+\ln ^2(2) . \\ &=\frac{1}{2} \ln ^2(2)\end{aligned}$$