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!

I am sure there are more tidy ways to do this but as an alternative...
Why not do $\int \ln(2x + 1)dx$ using:
$v^\prime = 1 \Rightarrow v = x$ and $u = \ln(2x+1)\Rightarrow u^\prime=\frac{2}{2x+1}$
Therefore, 
$$\int \ln(2x + 1)dx = x\ln(2x+1) - \int \frac{2x}{2x+1}dx$$
Then make the substitution $u=2x + 1$ to yield,
$$\int \ln(2x + 1)dx = x\ln(2x+1) - \int \frac{1}{2}\frac{u-1}{u}du$$
Thus,
$$\int \ln(2x + 1)dx = x\ln(2x+1) - \frac{1}{2}u - ln(u) + c$$
Then,
$$\int \ln(2x + 1)dx = x\ln(2x+1) - \frac{1}{2}(2x-1) - \ln(2x-1) + c$$
$$= \ln(2x-1)(x-1) - \frac{1}{2}(2x-1) + c$$
