Question

integrate ln(2x+1)dx

Answer 1

We did this once?

Answer 2

when?

Answer 3

maybe i was not here

Answer 4

Let p=2x+1
Then dp=2dx
Rewrite your integral as:
\[\frac{1}{2} \int\limits \ln(p)dp\]
Then use integration by parts:
u=ln(p) dv=dp
du=1/p v=p
so you have:
\[\frac{1}{2}p \ln(p)-\frac{1}{2} \int\limits \frac{p}{p}dp=\frac{1}{2}p \ln(p)-\frac{1}{2}p+C\]
Then replacing p you have:
\[\frac{1}{2}(2x+1)\ln(2x+1)-\frac{1}{2}(2x+1)+C\]