Iterated Integration with a fraction

In summary, the conversation is about a user struggling with iterated integration, specifically with the equation ∫ylnx/x from e^y to y for dx, y>0. The user mentions being out of practice with alternate methods for integration and not being able to solve the equation using u-substitution. Another user suggests a substitution of u=ln(x), which helps with the fraction in the equation. The original user realizes their mistake and thanks the other user for their help.
  • #1
I need help with Iterated Integration. I'm trying to integrate the following equation:

∫ylnx/x from e^y to y for dx, y>0

(sorry about the messy equation; first time user)

I know how to integrate ln, but the fraction has me thrown for a loop. I tried U-substitution, but the equation kept getting messier. To be clear, I'm out of practice with some of the alternate methods for integration, so I may not be considering something (I planned on practicing them over the summer, but I got a job and my time got eaten away).

Please help.
 
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  • #2
Did you try the rather obvious substitution: u= ln(x)?
 
  • #3
Yes; it's the fraction I'm having a problem with, though. I am not sure why, though. I've done them before. This semester though I haven't run into them often, and the ones I've seen were done in class as example problems (the teacher picks from his homework assignments).

I do have integration tables in the back of the book. I'll try those and see if one of them matches.
 
  • #4
Hello Millenniumf,

Welcome to PF.

The substitution suggested by HallsofIvy takes care of the "fraction" very nicely.

[itex]\displaystyle \int\frac{\ln(x)}{x}dx[/itex]

Let u=ln(x), then du = ?
 
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  • #5
Huh, it's obvious now! u=ln(x), du=(1/x), then I just input e^y and y and complete the problem.

I think my brain must have been fried from pulling an all-nighter getting my assignment completed; I've done more complex u-sub problems before with no complications.

I think I better start earlier in the day with future assignments.

Thank you for the help!
 

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