Power series of $f(x)=\frac{x+1}{x^2+x-4}$? I do not understand how I can find the power series with center in x=0 of the function
$$f(x)=\frac{x+1}{x^2+x-4}$$
I tried without succes in the following way:
$$f(x)=\frac{x+1}{x^2+x-4}=\frac{1}{2} \bigg[\frac{2x+1}{x^2+x-4}+\frac{1}{x^2+x-4}\bigg]$$
Now, for $\frac{2x+1}{x^2+x-4}$ I know that
$$\int \frac{2x+1}{x^2+x-4} dx=\mathrm{lg} (x^2+x-4) \implies \frac{2x+1}{x^2+x-4}=\sum_{n \geq 0} \frac{\mathrm{d}}{\mathrm{dx}} ((-1)^n \frac{(x^2+x-4)^{n+1}}{n+1})$$
The last one using the power series of logarithm.
And for $\frac{1}{x^2+x-4}$ I can complete the square and write
$$\frac{1}{x^2+x-4}=\frac{1}{(x-1/2)^2+15/4}=(4/15) \frac{1}{(\frac{x-1/2}{225/16})^2+1}=(4/15) \frac{1}{1-(-(\frac{x-1/2}{225/16})^2)}=(4/15) \sum_{n \geq 0} (-(\frac{x-1/2}{225/16})^2)^{n}$$
But in both cases I don't get a power series with center in $x=0$ (I'm not even sure that these two are really power series), so I'm wrong.
What is the proper way to write the power series of $f(x)$ with center in $x=0$?
 A: Hint
Just start rewriting $$f(x)=\frac{x+1}{x^2+x-4}=-\frac{1+x}{4-x-x^2}=-\frac 14 \frac {1+x}{1-\frac{x+x^2}4}$$ Now consider the development (by Taylor series) of $\frac 1{1-y}$ and then replace $y$ by $\frac{x+x^2}4$. 
When done, multiply by $-\frac{1+x}4$ and you should get something like $$f(x)=-\frac{1}{4}-\frac{5 x}{16}-\frac{9 x^2}{64}-\frac{29
   x^3}{256}+O\left(x^4\right)$$ which could have been obtained by division.
Edit
Another way to approach the problem.
From the very first terms obtained by division, you can set $$f(x)=-\sum_{i=0}^\infty \frac {a_i }{4^{i+1}}x^i$$ and write the problem as $$x+1=(x^2+x-4)f(x)\implies (1+x)+(x^2+x-4)\sum_{i=0}^\infty \frac {a_i }{4^{i+1}}x^i=0$$ Looking at $x^0$ and $x^1$, we immediately get $a_0=1$ and $a_1=5$. Now, for any $n\geq 2$, we have, just as  B. Goddard commented,$$a_n=a_{n-1}+4a_{n-2}$$ Using the standard method for recurrence relations, we have $$a_n=c_1 \left(\frac{1}{2}-\frac{\sqrt{17}}{2}\right)^n+c_2
   \left(\frac{1}{2}+\frac{\sqrt{17}}{2}\right)^n$$ and applying the initial conditions $a_0=1$ and $a_1=5$ $$a_n=\frac{1}{34} \left(\left(17-9 \sqrt{17}\right) \left(\frac{1}{2}
   \left(1-\sqrt{17}\right)\right)^n+\left(17+9 \sqrt{17}\right) \left(\frac{1}{2}
   \left(1+\sqrt{17}\right)\right)^n\right)$$ which are integer numbers which correspond to sequence $A006131$ in $OEIS$.
