<s>[INST] Find Maclaurin series of $\frac{x^2+3}{x^2-x-6}$ 
Find the Maclaurin series of $\frac{x^2+3}{x^2-x-6}$.

So far I have:
$$\frac{x^2+3}{x^2-x-6}=1+\frac{x+9}{(x-3)(x+2)}=1+\frac{x+2}{(x-3)(x+2)}+\frac{7}{(x-3)(x+2)}=1-\frac{1}{(3-x))}+\frac{7}{(x-3)(x+2)}$$
How should I continue?
 [/INST] Hint:
Set
$$
\frac{7}{(x-3)(x+2)} = \frac{A}{x-3} + \frac{B}{x+2}
$$
And solve for $A$ and $B$.
This is called a "partial fraction decomposition". To solve for $A$ and $B$, generally you multiply both sides by $(x-3)(x+2)$ and plug in values of $x$: $x = 3$ and $x = -2$ will work well.
Finally, you need to know how to find the McLaurin series of $\frac{c}{ax + b}$. To do so, note that
$$
\frac{c}{ax + b} = (c/b) \cdot \frac{1}{1 - \left[-\frac{a}{b} x\right]}
$$
and plug in the McLaurin series for $\frac{1}{1 - y}$.
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