Integration by parts - $\int \ln (2x+1) \text{dx}$ Use integration by parts to find
$\int \ln (2x+1) \text{dx}$.
So far I have:
$$x\ln(2x+1)-\int\dfrac{2x}{2x+1}dx+c$$
Using integration by substitution to find the integral
$$u=2x+1\Rightarrow\text{du}=2\text{dx}$$
$$\int\dfrac{2x}{2x+1}\cdot\dfrac{1}{2}\text{du}=\int xu^{-1}$$
$$=\int \left(\dfrac{u}{2}-\dfrac{1}{2}\right)u^{-1}\text{du}=\int\left[\dfrac{1}{2}-\dfrac{1}{2}u^{-1}\right]\text{du}$$
$$=\dfrac{1}{2}x-\dfrac{1}{2} \ln \left|2x+1\right|$$
Looking at the answer in the back, this is wrong.
The answer is $x \ln(2x+1)-x+\dfrac{1}{2}\ln(2x+1)+c$.
What have I done wrong?

Your issue is in what follows this step:
\begin{align*}
x\ln(2x+1) - \int \frac{2x}{2x+1}\,\text{dx} &= x\ln(2x+1) - \int \frac{2x+1-1}{2x+1}\,\text{dx}\\
&= x\ln(2x+1) - \left(\int \frac{2x+1}{2x+1}\,\text{d}x - \int \frac{1}{2x+1}\,\text{d}x\right)\\
&= x\ln(2x+1) - \left(\int \,\text{d}x - \int \frac{1}{2x+1}\,\text{d}x\right)\\
&= x\ln(2x+1) - \left(x - \frac{1}{2}\ln(2x+1)\right) + C\\
&= x\ln(2x+1) - x + \frac{1}{2}\ln(2x+1) + C
\end{align*}
(The second to last equality follows from a u-substitution, using $u = 2x+1$.  Technically, it seems that the answer should be
$$x\ln(2x+1)-x+\frac{1}{2}\ln|2x+1|+C$$
but perhaps there is some reason for them to be able to drop that absolute value.)
We can also compute this integral using substitution:  Use $u = 2x+1$, $2x = u-1$, $du = 2dx$, to compute it as follows:
\begin{align*}x\ln(2x+1) - \int \frac{2x}{2x+1}\,\text{d}x &= x\ln(2x+1) - \frac{1}{2}\int \frac{u-1}{u}\,\text{d}u\\
&= x\ln(2x+1) - \frac{1}{2}\int\,\text{d}u + \frac{1}{2}\int \frac{1}{u}\,\text{d}u\\
&= x\ln(2x+1) - \frac{u}{2} + \frac{1}{2}\ln|u| + C\\
&= x\ln(2x+1) - \frac{2x+1}{2} + \frac{1}{2}\ln|2x+1| + C\\
&= x\ln(2x+1) - x - \frac{1}{2} + \frac{1}{2}\ln|2x+1|+C\\
&= x\ln(2x+1) - x + \frac{1}{2}\ln|2x+1|+C'
\end{align*}
(the last equality is just a matter of absorbing the constant $\frac{1}{2}$ into the integration constant $C + \frac 12 = C'$)
