help solving an integral $\int\left(\frac {x-1}{3-x}\right)^\frac{1}{2} dx$ $\displaystyle\int \left(\frac {x-1}{3-x}\right)^\frac{1}{2}\,\rm dx$
I am stuck on this part:
Let $u=\dfrac{x-1}{3-x}~\longrightarrow$  $~~\rm du=\dfrac {2}{(x-3)^2}\,\rm dx,$ which can be represented as
$\rm du=\dfrac{1}{3-x} - \dfrac{1-x}{(3-x)^2}\,\rm dx$
I cannot "see" how to get to this $2$ $\displaystyle\int \: \frac{(u)^\frac{1}{2}}{(u+1)^2} \:\rm dx$
after this part I know how to solve it;
I just wish someone would show me "step by step" this part 
It seems it involves some sort of "leap" of thought; or is there a systematic way doing this using basic algebra?
Thanks.

First note
$$
du=\frac{2}{(x-3)^2}dx
$$
which implies 
$$
\frac{(x-3)^2}{2}du=dx.
$$
Also, 
$$
u+1=\frac{x-1}{3-x}+\frac{3-x}{3-x}=\frac{2}{3-x}
$$
which implies
$$
(u+1)^2=\left(\frac{2}{3-x}\right)^2=\frac{4}{(x-3)^2}.
$$
Putting it together you have
$$
\int \left(\frac {x-1}{3-x}\right)^\frac{1}{2} dx = \int u^{1/2}\cdot\frac{(x-3)^2}{2}du = 2\int u^{1/2}\cdot\frac{(x-3)^2}{4}du
= 2\int \frac{u^{1/2}}{(u+1)^2}du.
$$
