[SOLVED] Integration of a natural log

[krusty the clown]

I am asked to Integrate by parts

\int \ln(2x+1) dx

So,
\mbox{u}=ln(2x+1)
\mbox{du}=\frac{2}{2x+1}
\mbox{dv}=\mbox{dx}
\mbox{v}=\mbox{x}
I plug all of that in and I get,

{\int \ln(2x+1) dx\}={\mbox{x}\ln(2x+1)}-{\int \frac{2x}{2x+1} \mbox{dx}}

At this point I tried doing another substitution letting u equal 2x, but this just brought me back to the original question.
After this anything I try to do doesn't lead me anywhere. I know it can be solved because the answer is in the back of the book.
First, is what I have done so far correct?
Second, how do I finish the problem?
Any help you have would be greatly appreciated.

Erik

[jamesrc]

Try approaching it like this:

Let θ = 2x+1, so that your integral is:

\int{\ln\left(2x+1\right) dx} = 2\int{\ln\theta d\theta}

Now you can integrate by parts like you did before, with:

u = ln(θ) --> du = dθ/θ
dv = dθ --> v = θ

You should get:

2\left[ \theta\ln\theta - \theta \right]

as your result. Now all you have to do is plug θ = 2x+1 back in to find the final indefinite integral.

[HallsofIvy]

jamesrc's suggestion is good. If you want to go ahead do it the way you were, you can actually do the division \frac{2x}{2x+1}= 1- \frac{1}{2x+1}. Of course, to do the integral of \frac{1}{2x+1}, make the substition u= 2x+1 (which gets you right back to jamesrc's suggestion).

[krusty the clown]

thanks for your help, I hate it when the answer is so obvious and I can't figure it out.

Thanks, Erik

[krusty the clown]

One final question, when you substituted theda, shouldn't the constant in front of the integral be 1/2 instead of 2???

(using u instead of theda)

u=2x-1
du=2dx
dx=(1/2)du

[e(ho0n3]

Yes, it should be 1/2.

[jamesrc]

Yes, sorry about that. At least that proves that you know what you're doing.