Integrate: $\int\ln(2x+1) \, dx$ $$\int\ln(2x+1) \, dx$$
I setting up this problem and I am finding it hard to understand why $dv= 1$. 
When using this formula 
$$\int u\ dv=uv-\int v\ du$$
And using these Guidelines for Selecting $u$ and $dv$:
“L-I-A-T-E” Choose $u$ to be the function that comes first in this list:
L: Logrithmic Function
I: Inverse Trig Function A: Algebraic Function
T: Trig Function
E: Exponential Function 
A: \begin{align}
\int\underbrace{\ln(2x+1)}_\text{This is $u$.} \, dx & = \int u\,dx = ux - \int x\,du \\[10pt]
& = x\ln(2x+1) - \int x\cdot \frac{2}{2x+1} \,dx \quad\text{etc.} \\[10pt]
& = x\ln(2x+1) - \int \left( 1 - \frac 1 {2x+1} \right) \,dx \qquad \text{etc.}
\end{align}
A: This is a really tricky problem to see for the first time, but it does fit with the guidelines you give. 
The first choice if possible for $u$ is the logarithmic function. However, we also need another function in order to use integration by parts. What are we to do? Well, although it doesn't look like much, we can realize that $\ln(2x+1)=1\cdot\ln(2x+1)$, and now suddenly we have two functions and can try integration by parts. Turns out it works nicely enough too
$$\begin{align}\int\ln(2x+1)\,dx&=\int1\cdot\ln(2x+1)\,dx=x\ln(2x+1)-\int x\cdot\frac{2}{2x+1}\,dx
\end{align}$$
and it's not too hard to finish off the integral from here.
A: $$\int\ln(2x+1)\ \mathrm dx$$
Using integration by parts, we have
$$u=\ln(2x+1)\Rightarrow \mathrm du=\frac{2}{2x+1}\mathrm dx$$
$$\mathrm dv=\mathrm dx\Rightarrow v=x$$
Which yields
$$x\ln(2x+1)-\int\frac{2x}{2x+1}\mathrm dx$$
Using substitution, we have
$$s=2x+1\Rightarrow \frac12\mathrm ds=\mathrm dx$$
Therefore
$$x\ln(2x+1)-\frac12\int\frac{s-1}{s}\mathrm ds$$
$$=x\ln(2x+1)-\frac12\int\left(1-\frac{1}{s}\right)\ \mathrm ds$$
$$=x\ln(2x+1)-\frac12\left(\int\mathrm ds-\int\frac{1}{s}\mathrm ds\right)$$
$$=x\ln(2x+1)-\frac12\int\mathrm ds+\frac12\int\frac{1}{s}\mathrm ds$$
$$=x\ln(2x+1)-\frac12 s+\frac12\ln s+C$$
$$=x\ln(2x+1)-\frac12(2x+1)+\frac12\ln(2x+1)+C$$
$$=\left(x+\frac12\right)\ln(2x+1)-\left(x+\frac12\right)+C$$
$$=\left(x+\frac12\right)\big(\ln(2x+1)-1\big)+C$$
A: Another solution by substitution:
$u = 2x + 1$
$x = \frac{u-1}{2}$
$\frac{du}{dx} = \frac{1}{2}$
$\int \log(2x+1) dx = \frac{1}{2}\int \log(u) du = \frac{1}{2}(u \log(u) - u) + C = \frac{1}{2}((2x+1) log(2x+1) - (2x+1)) + C$
