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!

Here is a cute variant. Let $u=\ln(2x+1)$ and let $dv=dx$. Then $du=\frac{2}{2x+1}$ and (this is the cute part) we can take $v=x+\frac{1}{2}$. It follows that
$$\int \ln(2x+1)\,dx=\left(x+\frac{1}{2}\right)\ln(2x+1)-\int \left(x+\frac{1}{2}\right)\frac{2}{2x+1}\,dx.$$
But the remaining integrand is just $1$! It follows that our integral is
$$\left(x+\frac{1}{2}\right)\ln(2x+1) -x+C.$$
