I have the following integral: $$y=\int \frac{1}{1-2\sqrt{x}} \, dx$$
I first got $u=2\sqrt{x}$ which gives us $x=\frac{u^2}{4}$. Plugging this in I got: $$y=\int \frac{1}{1-2\sqrt{\frac{u^2}{4}}} \, du=\int \frac{1}{1-2(\frac{u}{2})} \, du=\int \frac{1}{1-u} \, du$$
After this I am not sure where to go from there. Even though there might be easier methods I'd prefer to stick with this method so please relate your responses to my correct or incorrect work shown above.
On
The substitution step that you made is incorrekt. Regarding the integral: $$\int{\frac{dx}{1-2\sqrt{x}}}$$ Let $x(\phi) = \phi^2/4$, then $\frac{dx}{d\phi}=\frac{\phi}{2}$, now substitute it: $$\frac{1}{2}\int{\frac{\phi}{1-\phi}d\phi}$$ To make it nicer to work with, you can rearrange it as follows: $$ \begin{align} \frac{1}{2}\int{\frac{1}{1-\phi}-1\ d\phi} &= -\frac{1}{2}\left(\log{|1-\phi|}+\phi\right)+K \\ &= -\frac{1}{2}\left(\log{|1-2\sqrt{x}|}+2\sqrt{x}\right)+K \end{align} $$ Where $K$ is an arbitrary constant.
We have $$ \int \frac{1}{1-2\sqrt{x}}dx$$ Substitute $u = 2\sqrt{x}$ so that $x = \frac{u^2}{4}$ and $dx = \frac{u}{2}du$. We get $$ \int \frac{u}{2(1-u)}du $$ Now, it can be done from here, but since you're learning, make another substitution which might help you seeing "typical" integral here.
Let $t = 1-u$ so that $u= 1-t$ and $du = -dt$. We get $$ \int \frac{1-t}{2t}(-dt) = -\frac{1}{2}\int \frac{1}{t} - \frac{t}{t} dt = -\frac{1}{2} \Big ( \int \frac{1}{t} dt - \int 1 dt \Big) $$
You should know those integrals. Preciselly: $$ \int \frac{1}{t} dt = \ln|t| + C_1 $$ and $$ \int 1 dt = t + C_2$$
So that taking $C=-\frac{1}{2}(C_1 + C_2) \in \mathbb R$ we have: $$ \int \frac{1}{1-2\sqrt{x}}dx = -\frac{1}{2} \big ( \ln|t| - t \big) + C $$ and now go back to variable $x$:
$$\int \frac{1}{1-2\sqrt{x}}dx= -\frac{1}{2}\ln|1-u| + \frac{1}{2}(1-u) + C = -\frac{1}{2}\ln|1-2\sqrt{x}| + \frac{1}{2}(1-2\sqrt{x}) + C $$