Integral with natural log problem

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Homework Help Overview

The problem involves the integral ∫(ln x)/(x + x ln x) dx, which presents challenges related to integration techniques, particularly integration by parts and substitution methods.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss various approaches to tackle the integral, including integration by parts and substitution methods. Some suggest changing variables to simplify the expression, while others share their attempts at manipulation, such as reaching the form ∫u/(1+u) du.

Discussion Status

The discussion is active, with participants sharing their thoughts and attempts. Some guidance has been offered regarding variable substitution, and there is a recognition of different interpretations of the problem. However, there is no explicit consensus on a single approach or solution yet.

Contextual Notes

Participants are navigating the complexities of the integral while learning integration techniques, and there is an acknowledgment of the use of integration tables by some members.

luxxx
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Hello,
The problem is ∫(ln x)/(x + x ln x) dx.

I've done most other problems in the set, but don't know where to start with this one. Although we are just learning integration by parts, I'm not sure how this would apply. I can get to ∫u/(1+u) du
Thanks for any help.

Homework Statement


Homework Equations


The Attempt at a Solution

 
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luxxx said:
Hello,
The problem is ∫(ln x)/(x + x ln x) dx.

I've done most other problems in the set, but don't know where to start with this one. Although we are just learning integration by parts, I'm not sure how this would apply. I can get to ∫u/(1+u) du
Thanks for any help.

When you have multiple versions of the exp and log function, try changing variables by letting w=e^x or w=ln(x) or x=ln(w) or x=e^w. Try those and see what happens.
 
I think the answer is 1 + ln x - ln ǀ1 + ln xǀ +C, but that's using an integration table.. I'd like to know how you would get there.
 
Well I did u substitution for u = ln x.
 
That would work. And note:

[tex]\frac{u}{1+u}=1-\frac{1}{1+u}[/tex]
 
Ha, yeah I just got that when you responded. Thank you!
 

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