An easier way to find the integral of: $\int {x\sqrt {2 + x} {\rm{ }}dx} $, where ${u^2} = 2 + x$ My attempt at the question:
$\eqalign{
  & \int {x\sqrt {2 + x} {\rm{ }}dx}   \cr 
  & {u^2} = 2 + x  \cr 
  & 2u{{du} \over {dx}} = 1  \cr 
  & {{du} \over {dx}} = {1 \over {2u}}  \cr 
  & u = \sqrt {2 + x}   \cr 
  & x = {u^2} - 2  \cr 
  & so:  \cr 
  & \int {x\sqrt {2 + x} {\rm{ }}dx}  = \int {x\sqrt {2 + x} } {\rm{ }}{{dx} \over {du}}du  \cr 
  &  = \int {x\sqrt {2 + x} } {\rm{ }} \times 2\sqrt {2 + x} du  \cr 
  &  = \int {2x} (2 + x)du  \cr 
  &  = \int {4x}  + 2{x^2}du  \cr 
  &  = \int {4({u^2} - 2)}  + 2{({u^2} - 2)^2}du  \cr  
  &  = \int {4{u^2} - 8}  + 2({u^4} - 4{u^2} + 4)du  \cr 
  &  = \int {2{u^4} - 4{u^2}} du  \cr 
  &  = {2 \over 5}{u^5} - {4 \over 3}{u^3} + C  \cr 
  &  = {2 \over 5}{(\sqrt {2 + x} )^5} - {4 \over 3}{(\sqrt {2 + x} )^3} + C  \cr 
  &  = {2 \over 5}{(2 + x)^{{5 \over 2}}} - {4 \over 3}{(2 + x)^{{3 \over 2}}} + C \cr} $

A few questions I have:
Given ${u^2} = 2 + x$, $u =  \pm \sqrt {2 + x} $, so why is it that we only take the principal square root and not the negative one for substitution?
The second question I have is a general one; is there an easier way of finding the integral? Have I done things in a manner that isn't overly longwinded? If so please suggest ways that would allow me to reach an answer quicker.
I'm on shakey grounds with integration at the moment so I was wondering if I could integrate this part of my working out without expanding out:
$ = \int {4({u^2} - 2)}  + 2{({u^2} - 2)^2}du$
Thank you for all your help!

I think it is simpler by parts:
$$u=x\;,\;\;u'=1\\v'=\sqrt{x+2}\;,\;\;v=\frac23(x+2)^{3/2}$$
so
$$\int x\sqrt{x+2}\,dx=\frac23x(x+2)^{3/2}-\frac23\int(x+2)^{3/2}dx=\frac23x(x+2)^{3/2}-\frac4{15}(x+2)^{5/2}+C=$$
$$=\frac2{15}(x+2)^{3/2}\left(3x-4\right)+C$$
