Proof of a dilogarithm identity

Through some experimentation, I've found:

$$\text{Li}_{2}(\sqrt{2}-1) \ + \text{Li}_{2}(1-\frac{1}{\sqrt{2}})\ =\frac{\pi^2}{8} - \frac{\ln^2(1 + \sqrt{2})}{2} - \frac{\ln^2{2}}{8} $$

I'm sure a simple proof is evading me, but I have not been able to prove this so far. I've been able to reduce it to proving the following integral equality: $$\frac{\ln^2{\sqrt{2}+1}}{4}-\frac{\pi^2}{16}=\int_{\sqrt{2}-1}^1 \frac{\ln{t}}{1-t^2}$$ Which, again, agrees numerically but seems to be as hard as proving the original problem. How can I prove this dilogarithm identity?

We use the standard notation $\operatorname{Li}_2$ for the dilogarithm function and $\zeta$ for the Riemann zeta function. For $|z|<1$, we have $$ \operatorname{Li}_2\left(2z-z^2\right) = -\ln^2\left(2-z\right)+\zeta(2)-2\operatorname{Li}_2\left(\frac{1}{2-z}\right)+2\operatorname{Li}_2(z). $$ This result can be obtained from Abel's identity by using appropriate transformations and substitutions. For further details see Lewin, pp. 7$-$11.

Let $z:=1-\sqrt2$ and divide both side by $2$. We have $$ \tfrac12\operatorname{Li}_2(-1) = -\tfrac12 \ln^2\left(1+\sqrt2\right)+\tfrac12\zeta(2) - \color{red}{\operatorname{Li}_2\left(\sqrt2 - 1\right)} + \operatorname{Li}_2\left(1-\sqrt2\right).\tag{$\spadesuit$} $$

From the duplication formula and the definition of the dilogarithm function, we have $$ \operatorname{Li}_2(-1) = -\tfrac12 \operatorname{Li}_2(1) = -\tfrac12 \zeta(2).\tag{$\clubsuit$} $$

Now if we express $\operatorname{Li}_2\left(1-\sqrt2\right)$ from $\left(\spadesuit\right)$ and substitute $\left(\clubsuit\right)$, we arrive at $$ \operatorname{Li}_2\left(1-\sqrt2\right) = \color{red}{\operatorname{Li}_2\left(\sqrt2 - 1\right)} + \color{blue}{\tfrac12 \ln^2\left(1+\sqrt2\right)-\tfrac34\zeta(2)}.\tag{$\heartsuit$} $$ Landen's identity states that for $z \notin \left(-\infty,0\right]$, we have $$ \operatorname{Li}_2\left(1-z\right) + \operatorname{Li}_2\left(1-\frac{1}{z}\right)=-\tfrac12\ln^2(z). $$

Let us substitute $z:=\sqrt2$ into Landen's identity. We arrive at $$ \operatorname{Li}_2\left(1-\sqrt2\right) + \color{green}{\operatorname{Li}_2\left(1-\frac{1}{\sqrt2}\right)}=\color{blue}{-\tfrac18\ln^2(2)}.\tag{$\diamondsuit$} $$ Now if we combine $\left(\heartsuit\right)$ and $\left(\diamondsuit\right)$ and use the fact that $\zeta(2) = \pi^2/6$, we arrive at $$ \color{red}{\operatorname{Li}_2\left(\sqrt2 - 1\right)} + \color{green}{\operatorname{Li}_2\left(1-\frac{1}{\sqrt2}\right)} = \color{blue}{\frac{\pi^2}{8} -\frac18\ln^2(2) - \frac12\ln^2\left(1+\sqrt2\right)}. $$

From this argument we can derive a general identity. For $|1-z|<1$ and $z \notin (-\infty,0]$, we have

$$ \operatorname{Li}_2\left(\frac{1}{z+1}\right)+\operatorname{Li}_2\left(1-\frac{1}{z}\right) = \tfrac12\left(\zeta(2)-\ln^2(z)-\ln^2(z+1)-\operatorname{Li}_2\left(1-z^2\right)\right). $$

For $z:=\sqrt2$, we arrive at your particular problem. We give another example. For $z:=1/\sqrt{2}$, we have $$ \operatorname{Li}_2\left(2-\sqrt2\right)+\operatorname{Li}_2\left(1-\sqrt{2}\right) = \frac{\pi^2}{24} - \frac38\ln^2(2)+\ln(2)\ln\left(\sqrt{2}+2\right)-\frac12\ln^2\left(\sqrt{2}+2\right). $$ At this example we used the special value $\operatorname{Li}_2\left(\frac12\right)=\frac{\pi^2}{12}-\frac12\ln^2(2)$.