Expert: Paul Klarreich - 9/10/2007

Question
Im doing integration by parts and im stock at this problem
The problem is Ln(2x+1)dx the formula is uv-Integration(vdu) my u= Ln(2x+1) du= (2/2x+1)
     my v= x        dv=dx

So i got the formula set and my mind whent blank and i cant get the secon part of the formula which im suppose to get ((2x+1)-1/(2x+1))dx i just can't figure it out right now that part i would really appriciate your help.

Answer
Questioner:   Israel
Category:  Calculus
Private:  No
 
Subject:  need some help to solve my problem
Question:  Im doing integration by parts and im stock at this problem
The problem is Ln(2x+1)dx the formula is uv-Integration(vdu) my u= Ln(2x+1) du= (2/2x+1)
    my v= x        dv=dx

So i got the formula set and my mind whent blank and i cant get the secon part of the formula which im suppose to get ((2x+1)-1/(2x+1))dx i just can't figure it out right now that part i would really appreciate your help.
....................................
Hi, Izzy,

Here is your integral:

{
| ln(2x + 1) dx
}

Integration by parts goes this way (as you already worked out):

u = ln(2x + 1),  du = 2/(2x + 1)

dv = dx,  v = x

So you have  (USE COURIER FONT TO VIEW THIS)
      
uv = x ln(2x + 1)   (one part)

and the other part is (actually there is a minus here, but I'll leave it out for now.)

{
| v du =
}

{   2x
| ------- dx =
} 2x + 1

OK, now -- how do we handle that fraction?  Once you have seen the trick, it is easy.  It amounts to something you learned in the fourth grade -- long division -- but that was so many years ago....

  2x
------ =
2x + 1

2x + 1 - 1   << see the trick of adding and subtracting 1?
----------- =
  2x + 1

2x + 1       1   
------- - ------- =
2x + 1    2x - 1

      1
1 - ------
   2x + 1

So that is your integral.  It is now easy to integrate.  The first term is just 1, which integrates to x, and the second term is a basic natural log integral.

I think you can handle the rest, including the minus in front of the  v du  integral.