$\displaystyle\int \left(\frac {x-1}{3-x}\right)^\frac{1}{2}\,\rm dx$
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.