Greetings I saw here (among the last integrals) that: $$\int_0^1 \ln\left(\frac{1-x}{1+x}\right)\ln\left(\frac{1-x^2}{1+x^2}\right)\frac{dx}{x}=\pi C$$ Where $C$ is Catalan's constant. Did this integral appear here before? (my quick search did not found anything).

I gave it a try and got stuck. Denoting the integral as $I$ and using that $\ln \left(\frac ab\right)=\ln a- \ln b\ $ we have: $$I=K(1,1)-K(1,-1)-K(-1,1)+K(-1,-1)$$ Where $$K(a,b)=\int_0^1\frac{\ln(1+ax)\ln(1+bx^2)}{x}dx$$ Differentiating under the integral sign: $$\frac{\partial^2}{\partial a \partial b}K(a,b)=\int_0^1 \frac{x^2}{(1+ax)(1+bx^2)}\,dx$$ By partial fractions we get: $$\frac{1}{a^2+b}\left(\int_0^1 \frac{ax}{bx^2+1}\,dx -\int_0^1 \frac{1}{bx^2+1}\,dx +\int_0^1 \frac{1}{ax+1} \,dx\right)$$ $$=\frac{1}{a^2+b}\left(\frac{a\ln(1+b)}{2b}-\frac{\arctan \left(\sqrt b\right)}{\sqrt{b}} +\frac{\ln(1+a)}{a}\right) $$ And now since $K(0,b)=K(a,0)=0$ $$K(a,b)=\frac12\int_0^a \int_0^b \frac{x\ln(1+y)}{y(x^2+y)}\,dy\,dx-\int_0^a \int_0^b \frac{\arctan \left(\sqrt y\right)}{\sqrt{y}(x^2+y)}\,dy\,dx +\int_0^a \int_0^b \frac{\ln(1+x)}{x(x^2+y)}\,dy\,dx$$ Is there a clever way to solve this? Another way is to start by using: $$-\frac12\ln\left(\frac{1-x}{1+x}\right)=\sum_{n=1}^\infty \frac{x^{2n+1}}{2n+1}$$ $$I=4\sum_{n,k=1}^\infty \frac{1}{(2n+1)(2k+1)}\int_0^1 x^{4n+2k+2}\,dx=4\sum_{n,k=1}^\infty \frac{1}{(2n+1)(2k+1)(4n+2k+3)}$$ But I dont know how to deal with this series. I would appreciate some help with this integral!