Integrate $\int_0^1\frac{\ln(1-x)\ln(x+x^2)}{1+x^2}dx $ I have been trying to derive
$$\int_0^1\frac{\ln(1-x)\ln(x+x^2)}{1+x^2}dx = \frac{\pi^3}{64} +\frac\pi{16}\ln^22-G\ln2$$
with $G$ being the the Catalan constant.
I noticed that a similarly-looking integral is posted and solved here. Although the solution is applicable and meritorious in itself, it seems an overkill to resort to the special function $\operatorname{Li}_3(z)$ given the elementary result.
 A: $$I=\int_{0}^{1}\frac{log(1-x)log(x+x^2)}{1+x^2}dx$$
$$=\underbrace{\int_{0}^{1}\frac{log(1-x)log(x)}{1+x^2}dx}_{I_1}+\underbrace{\int_{0}^{1}\frac{log(1-x)log(1+x)}{1+x^2}dx}_{I_2}$$
Let's start evaluating $I_2$ using Leibniz Rule:
$$I_2={\int_0^1}\frac{log(1-x)log(1+x)}{1+x^2}dx=-{\int_0^1}{\int_0^1}{\int_0^1}\frac{x^2}{(1+x^2)(1-yx)(1+xz)}dydzdx$$
$$={\int_0^1}{\int_0^1}{\int_0^1}dxdydz\\\left(\frac{(1+yz)+x(y-z)}{(1+y^2)(1+z^2)(1+x^2)}-\frac{y}{(1+y^2)(y+z)(1-yz)}-\frac{z}{(1+z^2)(y+z)(1+zx)}\right)$$
$$={\int_0^1}{\int_0^1}dydz\left(\frac{\frac{\pi}{4}(1+yz)+\frac{log(2)}{2}(y-z)}{(1+y^2)(1+z^2)}+\underbrace{\frac{log(1-y)}{(1+y^2)(y+z)}}_{y\rightarrow z\\ z\rightarrow y}+\frac{log(1+z)}{(1+z^2)(y+z)}\right)$$
$$={\int_0^1}{\int_0^1}dydz\left(\frac{\frac{\pi}{4}(1+yz)+\frac{log(2)}{2}(y-z)}{(1+y^2)(1+z^2)}+\frac{log\left(\frac{1-z}{1+z}\right)}{(1+z^2)(y+z)}\right)$$
$$=\int_0^1\left(\frac{\frac{\pi^2}{16}+\frac{log^2(2)}{2}}{1+z^2}+\underbrace{\frac{log\left(\frac{1-z}{1+z}\right)log\left(\frac{1+z}{z}\right)}{1+z^2}}_{z\rightarrow \frac{1-x}{1+x}}\right)dz$$
$$=\frac{\pi}{4}\left(\frac{\pi^2}{16}+\frac{log^2(2)}{4}\right)+log(2)\int_0^1\frac{log(x)}{1+x^2}dx-\underbrace{\int_{0}^{1}\frac{log(x)log(1-x)}{1+x^2}dx}_{I_1}$$
$$=\frac{\pi^3}{64}+\frac{\pi}{16}log^2(2)-Glog(2)-I_1$$
Hence:
$$I=I_1+I_2=\int_{0}^{1}\frac{log(1-x)log(x+x^2)}{1+x^2}dx
=\frac{\pi^3}{64}+\frac{\pi}{16}log^2(2)-Glog(2)\ \blacksquare$$
A: Write the numerator of the integrand as
$$\ln(1-x)\ln(x+x^2)=\ln(1-x)\ln x+\ln(1+x)\ln\frac{1-x}{1+x}+\ln^2(1+x)
$$
and, correspondingly
$$I=\int_0^1\frac{\ln(1-x)\ln(x+x^2)}{1+x^2}dx 
= I_1+I_2+I_3
$$
where
\begin{align}
I_1 &= \int_0^1\frac{\ln(1-x)\ln x}{1+x^2}dx 
= \frac12\int_0^1\frac{\ln^2(1-x)+\ln^2x-\ln^2\frac x{1-x}}{1+x^2}dx\\
I_2 &=\int_0^1\frac{\ln(1+x)\ln \frac{1-x}{1+x}}{1+x^2}dx 
\overset{ \frac{1-x}{1+x} \to x} = -\int_0^1\frac{\ln x\ln(1+x)}{1+x^2}dx-G\ln2 \\
I_3 &= \int_0^1\frac{\ln^2(1+x)}{1+x^2}dx 
= \int_0^\infty\frac{\ln^2(1+x)}{1+x^2}dx- \int_1^\infty \frac{\ln^2(1+x)}{1+x^2} \overset{x\to 1/x} {dx }\\
&= \int_0^\infty\frac{\ln^2(1+x)}{1+x^2}dx- \left(
I_3 + \int_0^1 \frac{\ln^2x}{1+x^2}dx-2 \int_0^1 \frac{\ln x\ln(1+x)}{1+x^2}dx\right)\\
&=\frac12 \int_0^\infty\frac{\ln^2(1+x)}{1+x^2}dx
 - \frac12 \int_0^1\frac{\ln^2 x}{1+x^2}dx
+\int_0^1 \frac{\ln x\ln(1+x)}{1+x^2}dx
\end{align}
Then, add up the three integrals above to obtain
\begin{align}
I=& \ I_1+I_2+I_3\\
=& \ \frac12 \int_0^1\overset{t=1-x}{\frac{\ln^2(1-x)}{1+x^2}}dx 
 +\frac12\int_0^\infty\overset{t=1+x}{ \frac{\ln^2(1+x)}{1+x^2}}dx 
 - \frac12\int_0^1\overset{t=x/(1-x)}{ \frac{\ln^2\frac x{1-x}}{1+x^2}}dx-G\ln2 \\
=& \ 2\int_0^\infty \underset{t^2\to 2t}  {\frac{t\ln^2t}{t^4+4}dt } -G\ln2 
=\frac{\pi^3}{64} +\frac\pi{16}\ln^22-G\ln2
\end{align}
