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Integrating x^3 lnx dx - am i on the right track?

  1. May 27, 2012 #1
    1. The problem statement, all variables and given/known data
    integrating x^3lnxdx


    2. Relevant equations



    3. The attempt at a solution

    i let u = lnx

    du/dx = 1/x
    xdu = dx
    x=e^u
    substituting that, i got e^(4u)udu

    then i let v = 4u
    dv/du = 4
    1/4du = dv

    substituting that, i got 1/4integral e^v vdv

    I haven't gone beyond that step yet. I was hoping someone more knowledgeable than I currently am could tell me if I am on the right track or let me know at what point I went wrong. Responses very much appreciated!
     
    Last edited: May 27, 2012
  2. jcsd
  3. May 27, 2012 #2
    You're fine up to this point! Using u-substitution won't do anything helpful for you at this point; you are basically left trying to integrate the same problem. I would suggest integration by parts!
     
  4. May 27, 2012 #3

    SammyS

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    Hello drjohnsonn. Welcome to PF !

    As scurty suggested, try integration by parts.
     
  5. May 27, 2012 #4
    Thanks! I am quite a novice as of now and have not quite gotten to that point yet. I suppose I'll have to wait a bit before I finish working on this. But is the reason for having to integrate by parts that there is no general product integration rule?
     
  6. May 27, 2012 #5
    Well, I immediately think integration by parts when I see a term that can be reduced by differentiating (u in this case) and a term that doesn't get "more complicated" by anti-differentiating (##e^{4u}## in this case). It's tough to explain, someone else might have a better general rule for when to integrate by parts.

    Also, for what it's worth, you could have integrated by parts right from the start with no u-substitution. ##u = ln(x), \quad dv = x^3 \ dx##
     
  7. May 27, 2012 #6

    SammyS

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    Correct.

    In fact integration by parts is based upon the product rule for differentiation.
     
  8. May 28, 2012 #7
    Thanks for the advice!
     
  9. May 28, 2012 #8
    Am I accurate in doing this:
    u = lnx dv=x^3dx
    du = 1/xdx v = Sx^3dx = x^4/4

    and since Sudv = uv - Svdu

    Sx^3lnxdx = 1/4(x^4)lnx - 1/4Sx^3dx = 1/4x^4lnx - 1/16x^4 + C

    if i skipped something, point it out. i kind of ended up looking at the answer before i finished and hope that didn't fool me into incorrectly solving it
     
  10. May 28, 2012 #9
    Looks good :smile:
     
  11. May 28, 2012 #10

    D H

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    There's an easy way to check: Differentiate. Differentiation is oftentimes far easier than integration (and the only time it isn't is when both are easy). Upon differentiation you should recover the original equation.

    It's always a good idea to double check your work.
     
  12. May 28, 2012 #11
    wootwoot
     
  13. May 28, 2012 #12
    Yes. I think you are on the right track. Now, you can do integration by parts again to get the answer.
     
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