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!

$$ = x\ln (2x + 1) + \ln |{(2x + 1)^{{1 \over 2}}}| - x + C  $$
$$ = \ln(2x + 1)^x +  \ln(2x + 1)^\dfrac 12 - x + C $$
$$= \ln({(2x + 1)^x \cdot(2x + 1)^\dfrac 12}) - x + C $$
$$= \ln{(2x + 1)^\dfrac{2x+1}{2} } - x + C $$
$$= \dfrac {1}{2}\cdot (2x+1) \ln{(2x + 1)} - x + C $$
