Integral from 0 to 16 of $\sqrt{x}/(x-4)$? $$\int_{0}^{16}\frac{\sqrt{x}}{x-4}dx$$
So I'm letting $u=\sqrt{x}$, $du=1/2\sqrt{x}$, $u^2=x$ and $dx=2\sqrt{x}du$.
I just don't really know what to do from here. I am trying different things and nothing is working.

As you say, there is a discontinuity, so let us just consider $\int_4^{16} \frac {\sqrt x}{x-4}dx$ to start. If we substitute $u=x-4$ this becomes $\int_0^{12}\frac{\sqrt {u+4}}{u}du$  Very near $u=0$ this is essentially $\int_0^\epsilon \frac 2u du$ which diverges logarthmically, so you can just report that the integral does not exist.  There is similar behavior on the other side of the discontinuity.  
