Dilogarithm inversion formula: $ \text{Li}_2(z) + \text{Li}_2(1/z) = -\zeta(2) - \log^2(-z)/2$

The classical proof (Lewin) of the 'inversion formula' $(4)$ is simply to use : $$\operatorname{Li}_2(z)=-\int\frac{\log(1-z)}z\,dz$$ and set $z:=-\frac 1x\,$ to get the derivative : $$\frac d{dx}\operatorname{Li}_2\left(-\frac 1x\right)=-\frac{\log\left(1+\frac 1x\right)}{-\frac 1x}\frac 1{x^2}=\frac{\log\left(1+\frac 1x\right)}x=\frac{\log(1+x)-\log(x)}x$$ Integrating this again gives (with the constant determined at $x=1$) : $$\operatorname{Li}_2\left(-\frac 1x\right)+\operatorname{Li}_2(-x)=2\,\operatorname{Li}_2(-1)-\frac12\log^2(x)$$

($\,x:=-z\,$ gives your relation of course)

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Concerning the 'Landen identity' $(3)$ the same method applies : \begin{align} \frac d{dx}\operatorname{Li}_2\left(\frac x{1+x}\right)&=-\frac{\log\left(1-\frac x{1+x}\right)}{\frac x{1+x}}\frac 1{(x+1)^2}=-\frac{\log\left(\frac 1{1+x}\right)}{x(x+1)}\\ &=\log(1+x)\left(\frac 1x-\frac 1{1+x}\right)\\ \end{align}

so that integrating again gives (with constant determined by $x=0$) : $$\operatorname{Li}_2\left(\frac x{1+x}\right)=-\operatorname{Li}_2(-x)-\frac 12\log^2(1+x)$$