$\int {x^{1/2}\over {x^{1/2}} - 3} dx$ The given problem is:
\begin{aligned} 
\int   {x^{1/2}\over {x^{1/2}} - 3} dx
\end{aligned}

The textbook (Larson, Edwards) 9th edition suggests to use u-substitution and let u be the denominator. 

So, I made $u =x^{1/2} -3$. So, $u+3 = x^{1/2}$ Thus, du is $\frac{1}{2\sqrt{x}}$. So, $2x^{1/2} du = dx$, and that follow $2(u+3) du = dx$.

So, I then put the original integral in terms of u:
\begin{aligned} 
2\int   {u+3{}\over {u}} du
\end{aligned}

When I carry out the integration, I do not get the correct answer. I noticed something was a bit fishy when I moved the u up and expanded to get $u^0 + 3u^{-1}$ which seemed a bit odd for this problem. I expect something to be wrong with my numerator when rewriting the function, but I cannot seem to find any errors.

Any help would be appreciated.
You have to substitute for the $\sqrt x-3$ in the denominator, the $\sqrt x$ in the numerator, and the $dx$. I think you only did two of the three.
Your problem appears to be in changing over from x to u.

$$\int \dfrac{x^\frac12}{x^\frac12-3}dx=2\int\dfrac{x}{x^\frac12-3}\left(\dfrac{x^{-\frac12}dx}2 \right)=$$

$$\int\dfrac{(u+3)^2}{u}du$$