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Homework Help: Simple Integral Help

  1. Sep 29, 2010 #1
    1. The problem statement, all variables and given/known data

    Evaluate the following integral:

    [integral](x^2)*ln(x)dx

    2. Relevant equations

    I believe this is substitution by parts...

    3. The attempt at a solution

    I chose u = ln(x), and dv = x^2. The problem I am having is I can't figure out what du and v should be, because I am thinking too much and confusing myself. If u = ln(x), then du should simply be 1/x, right? And if dv = x^2, then to get v, you just need the integral of x^2, which is just 2x. Is that right?

    If those are right, then how come I come up with the following answer, which I don't believe is right (and I will laugh if it is but I really don't think so):

    2x*lnx - 2x + C

    Thanks in advance.
     
    Last edited: Sep 29, 2010
  2. jcsd
  3. Sep 29, 2010 #2

    Dick

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    The INTEGRAL of x^2 isn't 2x.
     
  4. Sep 29, 2010 #3

    fss

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    No. You have differentiated, not integrated.
     
  5. Sep 29, 2010 #4
    Well that's what I was confused about.

    When you are setting up a substitution by parts problem...when you get u, you differentiate to get the du...and then you integrate to get v from dv, right?

    So it should be x^3/3, not 2x?
     
  6. Sep 29, 2010 #5

    fss

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    Correct. Technically speaking you'd have:

    u = ln x
    dv = x2 dx
     
  7. Sep 29, 2010 #6
    Okay, so I got du = 1/xdx, and v = x^3/3

    Then, I did: lnx * x^3/3 - [integral]x^3/3 * 1/xdx

    then: lnx * 1/3*x^3 - 1/3[integral]x^3/xdx

    which simplifies to: lnx * 1/3 * x^3 - 1/3[integral]x^2dx

    which is: lnx * 1/3 * x^3 - 1/3[integral]x^3/3 + C

    final answer: ln(x) * 1/3 * x^3 - 1/9 * x^3 + C

    Does that seem right?

    Thanks in advance for reading this all over...
     
  8. Sep 29, 2010 #7

    fss

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    Looks like you got it to me.
     
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