Need help understanding how the solution goes from one step to another $\int \frac{\sqrt{x+4}}{x}dx$

This is the integral and the solution has the following steps outlined

$$\int \frac{\sqrt{x+4}}{x}dx$$

$$u=\sqrt{x+4}$$ $$u^2=x+4$$ $$2u\,du=dx$$

$$\int \frac{u}{u^2-4}(2u\,du)$$ $$\int \frac{2u^2}{u^2-4}\,du$$

I'm very comfortable with doing all of the above... no issues there, but the next step is where I get lost:

$$\int \left(2+ \frac{8}{u^2-4}\right) \, du$$

It's probably something very small I'm overlooking, but how did they get the term $2$ and $8$ in the numerator of the other term?

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Long division. $2$ is the quotient and $8$ is the remainder. – Michael Hardy Oct 4 '13 at 4:01
The Maple command $$Student[Calculus1]:-IntTutor(sqrt(x+4)/x, x)$$ does the job. See that link for info. – user64494 Oct 4 '13 at 6:32

Isn't it just because $$\frac{2 u^2}{u^2-4} = 2+\frac{8}{u^2-4} = \frac{2(u^2-4)+8}{u^2-4} ?$$
Ohh... so its just a matter of substituting the variable $x$ and $u$ back in to get to the more simplified form? – free_mind Oct 4 '13 at 2:17
$$\frac{2u^2}{u^2-4} = 2\left[ \frac{u^2}{u^2-4} \right] = 2\left[ \frac{u^2-4+4}{u^2-4} \right] = 2\left[ 1 + \frac{4}{u^2-4} \right] = 2 + \frac{8}{u^2-4}$$