Find the integral of $\int {{2 \over {\sqrt x (x - 4)}}dx} $ given the substitution: $u = \sqrt x $ My stab at it:
$\eqalign{
  & \int {{2 \over {\sqrt x (x - 4)}}dx}   \cr 
  & u = \sqrt x   \cr 
  & {{du} \over {dx}} = {1 \over 2}{x^{ - {1 \over 2}}} = {1 \over {2\sqrt x }}  \cr 
  & {{dx} \over {du}} = 2\sqrt x   \cr 
  & \int {f(x)dx}  = \int {{2 \over {\sqrt x (x - 4)}} \times 2\sqrt x {\rm{ }}du}   \cr 
  &  = \int {{{4\sqrt x } \over {\sqrt x (x - 4)}}du}   \cr 
  &  = \int {{{4u} \over {u({u^2} - 4)}}du}   \cr 
  &  = \int {{4 \over {({u^2} - 4)}}du}   \cr 
  &  = \int {{4 \over {({u^2} - 4)}}du}   \cr 
  &  = 4 \times {1 \over {2u}} \times \ln |({u^2} - 4)| + C  \cr 
  &  = {{2\ln |(x - 4)|} \over {\sqrt x }} + C \cr} $

My problem is the answer is:
$\ln |{{\sqrt x  - 2} \over {\sqrt x  + 2}}| + C$
I dont know how to simplify to reach this, this is most likely because I've probably made a mistake integrating. I am aware that you can reproduce the above answer using partial fractions to simplify the expression that is in terms of U and then integrate, but i was wondering if it was possible to duplicate the answer with my way of doing things.. 
Thank you for your help!

You did just fine up to $\;\int \frac{4}{u^2-4}\,du\;\;$
Note that $$\int \frac{4}{u^2-4}du = \int \frac{4}{(u - 2)(u+2)}\,du = \int \left(\frac {A}{u - 2} + \frac{B}{u + 2} \right)\,du$$
That is, we can use partial fractions to break our integral into the sum of two nicely integrable functions by simply solving for the numerators $A, B$.
In this case, $A$ and $B$ turn out to be "nice": To solve for $A, B$, note that $A(u + 2) + B(u - 2) = 4$. 
Letting $u = 2 \implies  4A = 4 \implies A = 1$
Letting $u = -2 \implies -4B = 4 \implies B = -1$.
So our integral $$\int \left(\frac {A}{u - 2} + \frac{B}{u + 2} \right)\,du = \int \left(\frac 1{u-1} + \frac {-1}{u + 2} \right) \,du$$
Also note that for the solution you are given:
$$\ln \left|{{\sqrt x  - 2} \over {\sqrt x  + 2}}\right| + C = \ln\left|\sqrt{x} - 2\right| - \ln\left| \sqrt x + 2\right| + C$$
