Easy integration by parts help

Click For Summary

Homework Help Overview

The discussion revolves around the integration of the function ln(2x + 1) using integration by parts. Participants are exploring different substitution strategies and the implications of their choices.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster considers using U = 2x + 1 for substitution but is uncertain about the corresponding dV. Some participants suggest using ln(2x + 1) as u and 1 as dv/dx. Others propose a different substitution method involving p = 2x + 1, leading to a discussion about the simplicity of each approach and their equivalence.

Discussion Status

Participants have shared different methods for approaching the integral, with one participant expressing satisfaction after applying a suggested method. There is an ongoing exploration of the differences between the two substitution methods and their effectiveness.

Contextual Notes

Participants are navigating the complexities of choosing appropriate substitutions in integration by parts, with some uncertainty about the implications of their choices. The discussion reflects a focus on understanding the integration process rather than arriving at a definitive solution.

MillerGenuine
Messages
64
Reaction score
0

Homework Statement



integral of ln(2x +1)

Homework Equations


I know this is an easy problem but i can not seem to figure out what to substitute for my U and my dV. I was thinking on making my U= 2x+ 1. but then my problem is what would my dV be? ln U? lnx? the ln is throwing me off a bit. I am not sure what it is the ln of.


The Attempt at a Solution



U= 2x+1 dV= lnU (?)
dU/2= dx V= 1/U (?)
 
Physics news on Phys.org
When you want to integrate by parts, a good choice for u is that part which has a simple derivative, and for v that part which is easy to integrate. So you can choose ln(2x+1)=u and 1=dv/dx

ehild
 
ok perfect. got the answer. I have seen it done the way you just showed & i have also seen it done a different way. If we first substitute
p= 2x+1
dp/2 = dx
then once this is done we have 2 integral ln(p).
now from here we do integration by parts by choosing
u= ln(p) dV= dp
du= 1/p V= p

now is there any difference between the two? or any situations where one will work and the other wont?
 
Both methods have the same result, but substituting first 2x+1 by p is much simpler. Do not forget to plug in 2x+1 for p at the end.

ehild
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 21 ·
Replies
21
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
2
Views
2K
Replies
5
Views
2K