A double sum for the square of the natural logarithm of $2$. I am trying to show

\begin{eqnarray*}
\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{(n+m)^2 2^{n}} =(\ln(2))^2. 
\end{eqnarray*}

Motivation : I want to use this to calculate $ \operatorname{Li}_2(1/2)$. So I want a solution to the above that does not use any reference to dilogarithms and please avoid rational multiples of $\pi^2$ (if possible).
Right, lets turn it into a double integral. ( I know lots of you folks prefer integrals to plums.)
Show

\begin{eqnarray*}
\int_0^1 \int_0^1 \frac{xy \ dx  \ dy}{(1-xy)(2-xy)} =(\ln(2))^2. 
\end{eqnarray*}

Reassuringly Wolfy agrees
My try: Let $u=xy$, & the double integral becomes
\begin{eqnarray*}
\int_0^1 \frac{dy}{y}  \int_0^y \frac{u \ du }{(1-u)(2-u)} . 
\end{eqnarray*}
Partial fractions
\begin{eqnarray*}
 \frac{u}{(1-u)(2-u)} =\frac{1}{1-u} - \frac{2}{2-u}.
\end{eqnarray*}
Do the $u$ integrations to leave the $y$ integrals
\begin{eqnarray*}
-\int_0^1 \frac{\ln(1-y) dy}{y}  +2 \int_0^1 \frac{\ln(2-y) dy}{y}.
\end{eqnarray*}
The first integral is
\begin{eqnarray*}
-\int_0^1 \frac{\ln(1-y) dy}{y}  = \frac{ \pi^2}{6}.
\end{eqnarray*}
which I was hoping to avoid and even worse Wolfy says the second integral is divergent
So you have a choice of questions, where did I go wrong in the above ? OR how can we show the initially stated result?
 A: You only forgot to evaluate the second part of the $u$-integral at the lower limit:
\begin{align}
\int \limits_0^1 \int \limits_0^y \left[\frac{1}{1-u} - \frac{2}{2-u}\right] \mathrm{d} u \, \frac{\mathrm{d} y}{y} &= \int \limits_0^1 \frac{-\ln(1-y) + 2 \ln(2-y) \color{red}{-2\ln(2)}}{y} \mathrm{d} y \\
&= \int \limits_0^1 \frac{-\ln(1-y) + 2 \ln\left(1-\frac{y}{2}\right)}{y} \mathrm{d} y \\
&\hspace{-4.55pt}\stackrel{y = 2z}{=} \hspace{-2pt} \int \limits_0^{1/2} \frac{-\ln(1-y)}{y} \, \mathrm{d} y + \int \limits_{1/2}^1 \frac{-\ln(1-y)}{y} \, \mathrm{d} y + 2 \int \limits_0^{1/2} \frac{\ln(1-z)}{z} \, \mathrm{d} z \\
&\hspace{-8pt}\stackrel{y = 1-x}{=} \hspace{-2pt} \int \limits_0^{1/2} \frac{-\ln(x)}{1-x} \, \mathrm{d} x + \int \limits_0^{1/2} \frac{\ln(1-z)}{z} \, \mathrm{d} z \\
&= \int \limits_0^{1/2} \left[\frac{\mathrm{d}}{\mathrm{d} t} \ln(t)\ln(1-t)\right] \mathrm{d} t = \ln^2\left(\frac{1}{2}\right) = \ln^2(2) \, .
\end{align}
A: A solution without dilogarithm.
\begin{align}J&=\int_0^1 \int_0^1 \frac{xy}{(1-xy)(2-xy)}\,dx\,dy\\
&\overset{x=1-uv,y=\frac{1-u}{1-uv}}=\int_0^1 \int_0^1 \frac{1-u}{(1+u)(1-uv)}\,du\,dv\\
&=\int_0^1\int_0^1\left(\frac{2}{(1+u)(1+v)}-\frac{1-v}{(1+v)(1-uv)}\right)\,du\,dv\\
&=\int_0^1\int_0^1\frac{2}{(1+u)(1+v)}\,du\,dv-J\\
&=2\left(\int_0^1\frac{1}{1+u}\,du\right)^2-J\\
&=2\ln^2 2 -J\\
J&=\boxed{\ln^2 2}
\end{align}
NB:
$\displaystyle \,dx\,dy=\frac{u\,du\,dv}{1-uv}$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{\sum_{n = 1}^{\infty}\sum_{m = 1}^{\infty}
{1 \over \pars{n + m}^{2}\, 2^{n}} = \ln^{2}\pars{2}} \approx 0.4805:\ {\Large ?}}$.

\begin{align}
&\bbox[5px,#ffd]{\sum_{n = 1}^{\infty}\sum_{m = 1}^{\infty}
{1 \over \pars{n + m}^{2}\, 2^{n}}} =
\sum_{n = 1}^{\infty}\sum_{m = 1}^{\infty}
{1 \over 2^{n}}\
\overbrace{\bracks{-\int_{0}^{1}\ln\pars{x}x^{m + n - 1}\,\dd x}}
^{\ds{{1 \over \pars{n + m}^{2}}}}
\\[5mm] = &\
-\int_{0}^{1}\ln\pars{x}
\sum_{n = 1}^{\infty}\pars{x \over 2}^{n}
\sum_{m = 1}^{\infty}x^{m}\,{\dd x \over x} =
-\int_{0}^{1}\ln\pars{x}
{x/2 \over 1 - x/2}\,{x \over 1 - x}\,{\dd x \over x}
\\[5mm] = &\
-\int_{0}^{1}\ln\pars{x}
{x \over \pars{2 - x}\pars{1 - x}}\,\dd x =
2\int_{0}^{1}
{\ln\pars{x} \over 2 - x}\,\dd x - \int_{0}^{1}
{\ln\pars{x} \over 1 - x}\,\dd x
\\[5mm] = &\
2\int_{0}^{1/2}
{\ln\pars{2x} \over 1 - x}\,\dd x -
\int_{0}^{1}{\ln\pars{1 - x} \over x}\,\dd x
\\[5mm] \stackrel{\mrm{IBP}}{=}\,\,\,&
-2\int_{0}^{1/2}\mrm{Li}_{2}'\pars{x}\,\dd x +
\int_{0}^{1}\mrm{Li}_{2}'\pars{x}\,\dd x =
-2\,\mrm{Li}_{2}\pars{1 \over 2} + \mrm{Li}_{2}\pars{1}
\\[5mm] = &\
-2\
\underbrace{\bracks{{\pi^{2} \over 12} -
{1 \over 2}\ln^{2}\pars{2}}}_{\ds{\mrm{Li}_{2}\pars{1 \over 2}}}\ +\
\underbrace{\pi^{2} \over 6}
_{\ds{\mrm{Li}_{2}\pars{1}}} = \bbx{\ln^{2}\pars{2}} \\ &
\end{align}
A: We have
$$
\sum^{\infty}_{n=1}\frac{2^n-1}{2^n}\frac{x^n}{n^2}=\frac{x}{(1-x)(2-x)}\textrm{, }|x|<1.
$$
Hence
$$
I=\int^1_0\int^1_0\frac{xy}{(1-xy)(2-xy)}dxdy=\int^{1}_{0}\int^{1}_{0}\sum^{\infty}_{n=1}\left(1-2^{-n}\right)(xy)^ndxdy=
$$
$$
=\sum^{\infty}_{n=1}\left(1-2^{-n}\right)\left(\int^{1}_{0}x^ndx\right)^2=\sum^{\infty}_{n=0}\frac{1}{(n+1)^2}-2\sum^{\infty}_{n=0}\frac{1}{2^{n+1}(n+1)^2}.
$$
Hence using $\sum^{\infty}_{n=0}\frac{x^{n}}{n+1}=-\frac{\log(1-x)}{x}$, $|x|<1$, we get
$$
I=-2\int^{1}_{1/2}\frac{\log(1-x)}{x}dx+\int^{1}_{0}\frac{\log(1-x)}{x}dx=
$$
$$
=-\int^{1}_{1/2}\frac{\log(1-x)}{x}dx+\int^{1/2}_{0}\frac{\log(1-x)}{x}dx=
$$
$$
=-\int^{1/2}_{0}\frac{\log(x)}{1-x}dx
+\int^{1/2}_{0}\frac{\log(1-x)}{x}dx=\int^{1/2}_{0}\frac{d}{dx}\left[\log(1-x)\log(x)\right]dx=\log^22
$$
