How Can I Finish Integrating a Natural Log Using Integration by Parts?

[krusty the clown]

[SOLVED] Integration of a natural log

I am asked to Integrate by parts

[tex]\int \ln(2x+1) dx[/tex]

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

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

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:

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

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

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

You should get:

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

as your result. Now all you have to do is plug θ = 2x+1 back into 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 [itex]\frac{2x}{2x+1}= 1- \frac{1}{2x+1}[/itex]. Of course, to do the integral of [itex]\frac{1}{2x+1}[/itex], 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.