Find Maclaurin series of $\frac{x^2+3}{x^2-x-6}$

Question

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?

Answer 1

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 $\frac{1}{1 - y}$.

Answer 2

As in this answer, suppose $$ \frac{a+bx}{6+x-x^2}=\sum_{k=0}^\infty c_kx^k $$ then $$ \begin{align} a+bx &=\left(6+x-x^2\right)\sum_{k=0}^\infty c_kx^k\\ &=\sum_{k=0}^\infty c_k\left(6x^k+x^{k+1}-x^{k+2}\right)\\ &=\underbrace{\ \ \ \ 6c_0\ \ \ \ \vphantom{()}}_{a}+\underbrace{\left(6c_1+c_0\right)}_b\,x+\sum_{k=2}^\infty\left(6c_k+c_{k-1}-c_{k-2}\right)x^k \end{align} $$ Since the coefficient of $x^k$ is $0$ for $k\ge2$, we must have $6c_k+c_{k-1}-c_{k-2}=0$; that is, $$ c_0=\frac a6\quad c_1=\frac{6b-a}{36}\quad c_k=\frac{c_{k-2}-c_{k-1}}6 $$

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$$ \begin{align} \frac{x^2+3}{x^2-x-6} &=1+\frac{x+9}{x^2-x-6}\\ &=1+\frac{-9-x}{6+x-x^2}\\ &=1+\sum_{k=0}^\infty c_kx^k \end{align} $$ where $c_0=-\frac32$, $c_1=\frac1{12}$, and $c_k=\frac{c_{k-2}-c_{k-1}}6$.