How to write $\frac{x^3+4x}{x^2-4}$ as a serie at $x_0=1$? I did this, but what to do next?
$f(x)=x+\frac{4}{x-2}+\frac{4}{x+2}$

In this case the $n$-th derivative of $f$ is explicitly calculable:
$$f(x) = x + 4\,\frac{1}{x - 2} + 4\,\frac{1}{x + 2}$$
$$f'(x) = 1 - 4\,\frac{1}{(x - 2)^2} - 4\,\frac{1}{(x + 2)^2}$$
$$f''(x) =  4\,\frac{2}{(x - 2)^3} + 4\,\frac{2}{(x + 2)^3}$$
$$\cdots$$
$$f^{(n)}(x) =  (-1)^n4\,\frac{n!}{(x - 2)^{n+1}} + (-1)^n4\,\frac{n!}{(x + 2)^{n+1}}$$
Now, you can apply Taylor's theorem.
