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?
 A: Hint:
$$\ldots = \int\frac{\sqrt{x}-3+3}{\sqrt{x}-3}\,dx = \int \left(1-\frac{3}{\sqrt{x}-3}\right)\, dx \ldots$$
A: With $\sqrt x = u$
Let $\sqrt x = u$, then we have $x = u^2$ and $\mathrm{d}x=2u\,\mathrm{d}u$.
\begin{align}
\int\frac{u}{u-3}\times 2u\,\mathrm{d}u 
&= 2\int\frac{u^2}{u-3}\,\mathrm{d}u\\ 
&= 2\int\frac{u^2-9+9}{u-3}\,\mathrm{d}u\\
&= 2\int\frac{u^2-9}{u-3}\,\mathrm{d}u + 18\int\frac{1}{u-3}\,\mathrm{d}u\\ 
&= 2\int(u+3)\,\mathrm{d}u + 18\int\frac{1}{u-3}\,\mathrm{d}u\\ 
&= u^2 + 6 u+18\ln(u-3)+C_1\\
&= x+6\sqrt x +18\ln (\sqrt x -3) +C_1
\end{align}
With $\sqrt x - 3 = t$
Let $\sqrt x -3  = t$, then we have $x = (t+3)^2$ and $\mathrm{d}x=2(t+3)\,\mathrm{d}t$.
\begin{align}
\int\frac{t+3}{t}\times 2(t+3)\,\mathrm{d}t 
&= 2\int\frac{(t+3)^2}{t}\,\mathrm{d}t \\ 
&= 2\int\frac{t^2+6t+9}{t}\,\mathrm{d}t \\ 
&= \int\left(2t+12+\frac{18}{t}\right)\,\mathrm{d}t \\ 
&= t^2 + 12 t + 18 \ln t +C_2\\
&= (\sqrt x-3)^2 +12(\sqrt x -2) + 18\ln(\sqrt x- 3) +C_2\\
&= x +6\sqrt x +18\ln(\sqrt x -3) + C_2-15
\end{align}
Here $C_1=C_2-15$. You can choose either way and the difference is only in the constant.
With $\frac{\sqrt x}{\sqrt x -3}=v$
It is intentionally left for the readers as an exercise.
A: Let $\sqrt{x}=t$.
Thus, $dx=2tdt$ and
$$\int\frac{\sqrt{x}}{\sqrt{x}-3}dx=2\int\frac{t^2}{t-3}dt=2\int\frac{t^2-9+9}{t-3}dt=$$
$$=2\left(\frac{t^2}{2}+3t+9\ln|t-3|\right)+C=x+6\sqrt{x}+18\ln|\sqrt{x}-3|+C.$$
A: You can use the substitution $x=u^2$. So, the we differentiate both side with respect to $x$ and we get $dx=2udu$ and the the integrand becomes $\frac{2u^2}{u-3}du$. After we use it is easier to go along it.
A: if $t=\sqrt{x}-3$, $x=\left(t+3\right)^2$ then $\dfrac{dt}{dx}=\frac{1}{2\sqrt{x}}\implies dt=\frac{dx}{2\sqrt{x}}\implies2\sqrt{x}dt=dx$
so$$\frac{\sqrt{x}}{\sqrt{x}-3}dx=\frac{\sqrt{x}}{\sqrt{x}-3}d2\sqrt{x}dt=2\frac{x}{t}dt=2\frac{\left(t+3\right)^2}{t}dt=2\left[\frac{t^2+6t}{t}+\frac{9}{t}\right]dt=2\left[t+6+\frac{9}{t}\right]dt=\left[2t+12+\frac{18}{t}\right]dt$$
integrate this:$$\int2t+12+\frac{18}{t}dt=\int2tdt+\int12dt+\int\frac{18}{t}dt=t^2+12t+18\ln t+c{=\left(\sqrt{x}-3\right)^2+12\left(\sqrt{x}-3\right)+18\ln\left|\sqrt{3}-x\right|}{=x-6\sqrt{x}+9+12\sqrt{x}-36+18\ln\left|\sqrt{3}-x\right|}\\{=x+6\sqrt{x}+18\ln\left|\sqrt{3}-x\right|-27+c,c_1=-27+c}\\\rightarrow x+6\sqrt{x}+18\ln\left|\sqrt{3}-x\right|+c_1$$
