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

Let $\displaystyle u=\ln(2x+1)$ and $\displaystyle \,dv=\,dx$. Thus $\displaystyle \,du=\frac{2\,dx}{2x+1}$ and $\displaystyle v=x$.

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

To integrate $\displaystyle \int\frac{2x\,dx}{2x+1}$, let $\displaystyle u=2x+1\implies x=\frac{u-1}{2}$

Thus, $\displaystyle \,du=2\,dx$

The integral is transformed into $\displaystyle \frac{1}{2}\int\frac{u-1}{u}\,du=\frac{1}{2}\int\left[1-\frac{1}{u}\right]\,du=\frac{1}{2}u-\frac{1}{2}\ln|u|+C$

Resubstitute:

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

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

$\displaystyle \implies \left(x+\frac{1}{2}\right)\ln(2x+1)-\frac{1}{2}(2x+1)+C$

$\displaystyle \implies \frac{1}{2}(2x+1)\ln(2x+1)-x\underbrace{-\frac{1}{2}+C}_{C}$

$\displaystyle \implies \color{red}\boxed{\frac{1}{2}(2x+1)\ln(2x+1)-x+C}$

The $\displaystyle \frac{1}{2}$ and $\displaystyle C$ make up another constant, which I called $\displaystyle C$...

I hope that this clarifies things!

--Chris