Integrate by parts: $\int \ln (2x + 1) \, dx$ $$\eqalign{
  & \int \ln (2x + 1) \, dx  \cr 
  & u = \ln (2x + 1)  \cr 
  & v = x  \cr 
  & {du \over dx} = {2 \over 2x + 1}  \cr 
  & {dv \over dx} = 1  \cr 
  & \int \ln (2x + 1) \, dx = x\ln (2x + 1) - \int {2x \over 2x + 1} \cr 
  &  = x\ln (2x + 1) - \int 1 - {1 \over {2x + 1}}   \cr 
  &  = x\ln (2x + 1) - (x - {1 \over 2}\ln |2x + 1|)  \cr 
  &  = x\ln (2x + 1) + \ln |(2x + 1)^{1 \over 2}| - x + C  \cr 
  &  = x\ln (2x + 1)^{3 \over 2} - x + C \cr} $$

The answer $ = {1 \over 2}(2x + 1)\ln (2x + 1) - x + C$
Where did I go wrong?
Thanks!

Starting from your second to last line (your integration was fine, minus a few $dx$'s in you integrals):
$$ = x\ln (2x + 1) + \ln |{(2x + 1)^{{1 \over 2}}}| - x + C  \tag{1}$$
Good, up to this point... $\uparrow$. 
So the error was in your last equality at the very end:
You made an error by ignoring the fact that the first term with $\ln(2x+1)$ as a factor also has $x$ as a factor, so we cannot multiply the arguments of $\ln$ to get $\ln(2x+1)^{3/2}$. What you could have done was first express $x\ln(2x+1) = \ln(2x+1)^x$ and then proceed as you did in your answer, but your result will then agree with your text's solution.
Alternatively, we can factor out like terms.
$$ = x\ln(2x + 1) + \frac 12 \ln(2x + 1) - x + C \tag{1}$$
$$= \color{blue}{\bf \frac 12 }{\cdot \bf 2x} \color{blue}{\bf \ln(2x+1)} + \color{blue}{\bf \frac 12 \ln(2x+1)}\cdot {\bf 1} - x + C$$
Factoring out $\color{blue}{\bf \frac 12 \ln(2x + 1)}$ gives us
$$= \left(\dfrac 12\ln(2x + 1)\right)\cdot \left(2x +1\right) - x + C $$ $$= \frac 12(2x + 1)\ln(2x+1) - x + C$$
