I don't know how to solve this integral 
Could you please help me compute the following integral: $$\int \frac{(x^2+x-4)}{x(x^2+4)}dx$$

What I've done so far is:

$$\int \frac{(x^2+x-4)}{x(x^2+4)}dx = \int \frac{A}{x}dx+\int \frac{B}{x^2+4}dx$$

So

$$x^2+x-4 = A(x^2-4)+Bx$$

And I did this so I could figure out the value of A and B but I am having a rough time trying to calculate A because of the $x^2$.
I've also seen that the solution solves this exercise by separating the fraction's numerator like this:

$$\int \frac{(x^2+x-4)}{x(x^2+4)}dx = \int \frac{x^2}{x(x^2+4)}dx+\int \frac{x}{x(x^2+4)}dx + \int \frac{-4}{x(x^2+4)}dx $$

When do I know that I have to use this? Is what I though correct? If yes, how do I continue it?
Thank you very much. Please, let me know if something is not very clear in my question.
Agapita.

To make the finding of partial fractions slightly easier you may care to write
$$\frac{x^2 + x - 4}{x(x^2 + 4)} = \frac{(x^2 + 4) + x - 8}{x(x^2 + 4)} = \frac{1}{x} + \frac{1}{x^2 + 4} - \frac{8}{x(x^2 + 4)},$$
and then it is prehaps easier to guess/see that
$$\frac{8}{x(x^2 + 4)} = \frac{2}{x} + \frac{-2x}{x^2 + 4}.$$
