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**stevedave** I'm having some trouble with this

$\displaystyle \int{\ln{(2x+1)}dx}$

I set $\displaystyle u = ln(2x+1)$ and $\displaystyle dv = dx$ so that $\displaystyle du=\frac{2}{2x+1}$ and $\displaystyle v = x$.

so after integrating by part once I get:

$\displaystyle x\ln{(2x+1)}-2\int{\frac{x}{2x+1}dx}$

then I tried to integrate by part again with $\displaystyle u=x$ and $\displaystyle dv=(2x+1)^{-1}$

my final answer ends up as $\displaystyle x\ln{(2x+1)}$ but my solution manual tells me differently.

Can anyone tell me what I'm doing wrong?