separating equation $$
\begin{align}
\frac {dy}{dx} &= \frac{(y-1)(x+2)}{(y+1)(x-3)}\\
\frac{y+1}{y-1}dy &= \frac{x+2}{x-3}dx
\end{align}
$$
Integrate both sides :
$$
\begin{align}
\int\frac{y+1}{y-1}dy & =\int\frac{x+2}{x-3}dx\\
\int 1+\frac{2}{y-1}dy &=\int 1+\frac{5}{x-3}dx\\
y+2\ln|y-1|&=x+5\ln|x-3|+C\\
\end{align}
$$
trapped here because I cannot get rid of $2\ln|y-1|$, any techniques I should use to solve this?

One first thing, you should put the absolute values there. The correct way to write down the solution is
$$y  + 2\ln \left| {y - 1} \right| = x + 4\ln \left| {x - 3} \right| + C$$
At this situation, they say that the ordinary differential equation is solved. However, there is not always a way to explicitly find $y$ in terms of $x$ from a relation of the form $h(x,y)=0$. In your case, it seems to happen. If you want to just get rid of $\ln \left| {y - 1} \right|$, do the following
$$\eqalign{
  & \ln {\left| {y - 1} \right|^2} - \ln {\left| {x - 3} \right|^4} =  - y + x + C  \cr 
  &   \cr 
  & \ln \left( {\frac{{{{\left| {y - 1} \right|}^2}}}{{{{\left| {x - 3} \right|}^4}}}} \right) =  - y + x + C  \cr 
  &   \cr 
  & \frac{{{{\left| {y - 1} \right|}^2}}}{{{{\left| {x - 3} \right|}^4}}} = {e^{ - y + x + C}}  \cr 
  &   \cr 
  & {\left| {y - 1} \right|^2}{e^y} = {e^C}{\left| {x - 3} \right|^4}{e^x} \cr} $$
