Expressing partial fractions in ascending powers of $x$ Let $f(x) = \frac {5x^2 + x + 6}{(3-2x)(x^2 + 4)}$
$i) $ Express $f(x)$ in partial fractions.
$ii)$ Hence obtain the expansion of $f(x)$ in ascending powers of $x$, up to and including the term in $x^2$.
I got part $i)$ but how do I do part $ii)$, I know how to the binomial expansion of each individual component, but I don't get the correct answer.
For $i)$ I got this, which is correct:
$f(x) = \frac {3}{3 - 2x} + \frac {-x -2}{x^2 + 4}$

Note that we can expand $\frac{3}{3-2x}$ as a geometric series given by
$$\begin{align}
\frac{3}{3-2x}&=\frac{1}{1-(2/3)x}\\\\
&=\sum_{n=0}^\infty \left(\frac23\right)^n\,x^n\\\\
&=1+\frac23x+\frac49x^2+O(x^3)
\end{align}$$
for $|x|<3/2$.
Analogously, we can expand $\frac{-x-2}{x^2+4}$ as a geometric series given by
$$\begin{align}
\frac{-(x+2)}{x^2+4}&=\frac{-(x+2)/4}{1+(x/2)^2}\\\\
&=-\frac14(x+2)\sum_{n=0}^\infty \left(\frac{-1}{4}\right)^n x^{2n}\\\\
&=-\frac12-\frac14x+\frac18x^2+O(x^3)
\end{align}$$
for $|x|<1$.
Therefore for $|x|<1$, we find that 
$$\bbox[5px,border:2px solid #C0A000]{\frac{5x^2+x+6}{(3-2x)(x^2+4)}=1+\frac5{12}x+\frac{41}{72}x^2+O(x^3)}$$
