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?

You made a mistake when evaluating the last integral. One approach to evaluate the  integral of $\tfrac{2x}{2x+1}$ is to write it as: $$\eqalign{\int\dfrac{2x}{2x+1}\mathrm dx&=\int\dfrac{-1+(2x+1)}{2x+1}\mathrm dx\\&=\int\left[\dfrac{-1}{2x+1}+\dfrac{2x+1}{2x+1}\right]\mathrm dx\\
&=\int\left[\dfrac{-1}{2x+1}+1\right]\mathrm dx\\
&=\int\dfrac{-1}{2x+1}\mathrm dx+\int1\,\mathrm dx\\
&=\int-\dfrac{1}{2x+1}\mathrm dx+x.\\
&=-\int\dfrac{1}{2x+1}\mathrm dx+x.\\
}$$
To evaluate the remaining integral use the substitution $u=2x+1$, then $\mathrm du=2\,\mathrm dx\ldots$ 
