How do I solve $\int\frac{\sqrt{x}}{\sqrt{x}-3}\,dx?$ $$
\int \frac{\sqrt{x}}{\sqrt{x}-3}dx
$$
What is the most dead simple way to do this?

My professor showed us a trick for problems like this which I was able to use for the following simple example:
$$
\int \frac{1}{1+\sqrt{2x}}dx
$$
Substituting:
$u-1=\sqrt{x}$ 
being used to create 
$\int\frac{u-1}{u}$
which simplifies to the answer which is:
$1+\sqrt{2x}-\ln|1+\sqrt{2x}|+C$
Can I use a similar process for the first problem?

Hint:
$$\ldots = \int\frac{\sqrt{x}-3+3}{\sqrt{x}-3}\,dx = \int \left(1-\frac{3}{\sqrt{x}-3}\right)\, dx \ldots$$
