Computing $ \sum_{n=1}^{\infty} \frac{1}{\phi^{2n}n^2}$ I would like to show that
$$ \sum_{n=1}^{\infty} \frac{1}{\phi^{2n}n^2}=\frac{\pi^2}{15}-\ln^2(\phi) $$
where $$ \phi = \frac{1+\sqrt{5}}{2}$$
How can this result be derived from the integral representation
$$  \sum_{n=1}^{\infty} \frac{1}{\phi^{2n}n^2} = -\int_0^{1/\phi^2}\frac{\ln(1-x)}{x} \mathrm dx $$
?

From the identity
$$ Li_2(x)+Li_2(1-x)=\frac{\pi^2}{6}-(\log x)(\log(1-x)) $$
it follows:
$$ Li_2(\phi^{-1})+Li_2(\phi^{-2}) = \frac{\pi^2}{6} - 2\log^2\phi. $$
From the identity
$$ Li_2(1-x)+Li_2(1-x^{-1})=-\frac{1}{2}\log^2 x$$
it follows:
$$ Li_2(-\phi^{-1})+Li_2(\phi^{-2}) = -\frac{1}{2}\log^2\phi.$$
From the identity
$$ Li_2(x)+Li_2(-x) = \frac{1}{2} Li_2(x^2)$$
and the previous ones it follows that:
$$ \frac{5}{2}Li_2(\phi^{-2}) = \frac{\pi^2}{6}-\frac{5}{2}\log^2\phi, $$
so:
$$ Li_2(\phi^{-2}) = \frac{\pi^2}{15}-\log^2(\phi). $$
To prove the used identities, it is sufficient to derive them with respect to $x$ and consider that:
$$ \frac{d}{dx}Li_2(x) = \frac{-\log(1-x)}{x},\quad \frac{d}{dx}\log^2{x}=\frac{2\log x}{x},\quad \frac{d}{dx}(\log x \log(1-x)) = \frac{\log(1-x)}{x}-\frac{\log x}{1-x},\quad Li_2(0)=0,\quad Li_2(1)=\zeta(2)=\frac{\pi^2}{6}.$$
