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?

$$\frac{2u^2}{u^2-4} = 2\left[ \frac{u^2}{u^2-4} \right] = 2\left[ \frac{u^2-4+4}{u^2-4} \right] = 2\left[ 1 + \frac{4}{u^2-4} \right] = 2 + \frac{8}{u^2-4}$$
