<s>[INST] 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.
 [/INST] Your partial fraction decomposition should be
$$\frac{x^2+x-4}{x(x^2+4)} = \frac{A}{x}+\frac{Bx+C}{x^2+4}. \tag{1}$$
Multiplying both sides by $x(x^2+4)$ results in
$$x^2+x-4 = (A+B)x^2+Cx+4A. \tag{2}$$
Consequently, $4A=-4 \implies A=-1$, $C=1$ and $A+B = 1 \implies B =2$.
After partial-fraction decomposition, the integral can be computed using standard transformations.
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