Finding the 11th term of $\frac{7x^2+2x+6}{(x+2)(x^2+1)}$ Taylor expansion 
Given $f(x)=\frac{7x^2+2x+6}{(x+2)(x^2+1)}$, find $f^{(11)}(0).$

I understood that we first need to use partial fractions to simplify the function.
$$\frac{7x^2+2x+6}{(x+2)(x^2+1)}=\frac{A}{(x+2)}+\frac{Cx+B}{(x^2+1)}=\frac{6}{(x+2)}+\frac{x}{(x^2+1)}$$
That can be written as $$\frac{6}{(x+2)}+\frac{x}{(x^2+1)}=\frac{1}{2}*\frac{6}{(1-(-\frac{x}{2}))}+\frac{x}{(1-(-x^2))}=\frac{3}{(1-(-\frac{x}{2}))}+\frac{x}{(1-(-x^2))}$$
The taylor expansion of $\frac{1}{1-x}=1+x+x^2+...+x^{11}+o(x^{12})$
So we have $$\frac{3}{(1-(-\frac{x}{2}))}+\frac{x}{(1-(-x^2))}$$
Taking the 11th derivative and plugin x=0 will give $$\frac{3}{\frac{11!}{2^11}}+\frac{x}{0}$$ and that can not be, where did I get it wrong?

We have
$$ f(x)=\frac{7x^2+2x+6}{(x+2)(x^2+1)}=\frac{6}{x+2}+\frac{x}{1+x^2}\tag{1}$$
through the residue theorem, then:
$$ f(x) = 3\sum_{n\geq 0}\frac{(-1)^n}{2^n}x^{n}+\sum_{n\geq 0}(-1)^n x^{2n+1} \tag{2}$$
by exploiting geometric series. It follows that:
$$ [x^{11}]\,f(x) = -\frac{3}{2^{11}}-1 \tag{3} $$
so:
$$ f^{(11)}(0) = \color{red}{- 11!\cdot\left(1+\frac{3}{2^{11}}\right)}.\tag{4}$$
