Recurrence relation with generating function problem I've got a recurrence problem that I'm close to solving, but having trouble with finishing up.

Solve the following recurrence relation using generating functions: 
  $$g_n = g_{n-1} + g_{n-2} + n$$
  for $ n>=2, g_0 = 1, g_1 =2 $.

What I've done so far:
$$G(x) = \sum_{n=0}^{\infty}g_nx^n$$
$$g_n < 0 = 0$$
Changed $g_n$ to work for all $n >= 0$:
$$g_0 = 1 = g_{n-1} + g_{n-2} + n + [n=0] $$
$[n=0]$ is conditional on whether n is 0. This makes $g_n$ work for $g_0$ and $g_1$.
I want it in the form of $G(x)$, so I multiply by $x^n$ and sum over $x>=0$:
$$\sum_{n=0}^{\infty}g_nx^n = \sum_{n=0}^{\infty}g_{n-1}x^n +\sum_{n=0}^{\infty}g_{n-2}x^n + \sum_{n=0}^{\infty}nx^n + \sum_{n=0}^{\infty}[n=0]x^n$$
Factor $x$ out of the first term on the right, and $x^2$ from the second. Third is rewritten, and the final term is just 1.
$$G(x) = xG(x) +x^2G(x)+ \frac{x}{(x-1)^2}+1 $$
Solving for $G(x)$:
$$G(x) = \frac{-x^2+x-1}{(x-1)^2(x^2+x-1)}$$
This is the point at which I'm stuck. From what I've understood from lecture, I want to do a partial fraction decomposition with the parts in the form $1/(1-x)$, so they can be expressed as $\sum_{n=0}^{\infty}px^n$ so that I can extract the coefficients to create the closed formula. Once I have them in that form, I know how to proceed. I've gotten this PFD, but I can't seem to figure out how to get it in the right form:
$$\frac{-2x-4}{x^2+x-1}+\frac{2}{x-1}-\frac{1}{(x-1)^2}$$
So either I've messed up somewhere in the problem, or I'm missing something on how to get this in the right form. Thanks for any suggestions.

A simpler way to set this kind of problems up is to write the recurrence with no subtraction in indices:
$$
g_{n + 2} = g_{n + 1} + g_n + n + 2
$$
Multiply by $x^n$, sum over $n \ge 0$, and recognize:
\begin{align}
\sum_{n \ge 0} a_{n + k} x^k
  &= \frac{G(x) - g_0 - g_1 z - \ldots - g_{k - 1} x^{k - 1}}{x^k} \\
\sum_{n \ge 0} x^n
  &= \frac{1}{1 - x} \\
\sum_{n \ge 0} n x^n
  &= x \frac{\mathrm{d}}{\mathrm{d} x} \frac{1}{1 - x} 
\end{align}
and get:
$$
\frac{G(x) - 1 - 2 x}{x^2}
  = \frac{G(x) - 1}{x} + G(x) + \frac{x}{(1 - x)^2} + 2 \frac{1}{1 - x}
$$
which gives:
\begin{align}
G(x)
 &= \frac{1 - x + x^2}{(1 - x)^2 (1 - x - x^2)} \\
 &= \frac{4 + 2 x}{1 - x - x^2} - \frac{2}{1 - x} - \frac{1}{(1 - x)^2}
\end{align}
Mow it comes handy to know that for the Fibonacci numbers $F_n$, defined by $F_0 = 0$, $F_1 = 1$, $F_{n + 2} = F_{n + 1} + F_n$ you have:
\begin{align}
\sum_{n \ge 0} F_n x^n
  &= \frac{x}{1 - x - x^2} \\
\sum_{n \ge 0} F_{n + 1} x^n
  &= \frac{1}{1 - x - x^2}
\end{align}
and using:
$$
\binom{-2}{n} = (-1)^n \binom{n + 2 - 1}{2 - 1}
              = (-1)^n (n + 1)
$$
you get:
\begin{align}
g_n
  &= 4 F_{n + 1} + 2 F_n - 2 - (n + 1) \\
  &= 2 F_{n + 2} + 2 F_{n + 1} - n - 3 \\
  &= 2 F_{n + 3} - n - 3
\end{align}
