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Integration by parts

  1. Dec 20, 2007 #1
    little help? simple question...

    1. The problem statement, all variables and given/known data


    Integrate:
    xlnx dx

    use integration by parts


    2. Relevant equations



    3. The attempt at a solution

    let u = x du = dx
    dv = ln x v = ??
     
  2. jcsd
  3. Dec 20, 2007 #2

    rock.freak667

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    If you don't know the integral of lnx...try letting u=lnx and dx=x dx instead of the way you wrote it
     
  4. Dec 20, 2007 #3
    1. The problem statement, all variables and given/known data


    Integrate:
    xlnx dx



    2. Relevant equations



    3. The attempt at a solution

    let u = x du = dx
    dv = ln x v = ??
     
  5. Dec 20, 2007 #4

    nicksauce

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    How about

    Let u = lnx
    du = 1/x
    dv = xdx
    v = x^2 / 2
     
  6. Dec 20, 2007 #5
    if I have to integrate it from 1 to 2 where are these plugged in?
     
  7. Dec 20, 2007 #6

    nicksauce

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    [tex]\int_a^b udv = uv|_a^b - \int_a^b vdu[/tex]
    Or in other words
    [tex]\int_a^b udv = u(b)v(b) - u(a)v(a) - \int_a^b vdu[/tex]
     
  8. Dec 20, 2007 #7
    my attempt

    i wonder if this worked...

    [tex]\int_1^2 x lnx dx = lnx (x^2/2)|_1^2 - \int_1^2 (x^2/2)* (1/x) dx [/tex]
    [tex] = 2ln2 - ((1/4)(x^2))|_1^2 [/tex]

    = 2ln2 - (5/4)
    ??
     
  9. Dec 20, 2007 #8

    HallsofIvy

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    Almost! [itex](1/4)(x^2)[/itex] when x= 2 is 1. when x= 1, it is 1/4 so the last part is -(1- 1/4)= -3/4, not -5/4.
     
  10. Dec 20, 2007 #9
    check your evaluation for the last part

    it might be me but you did the integration correctly, gj.
     
  11. Dec 20, 2007 #10

    nicksauce

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    Almost...

    [tex]x^2|^2_1[/tex] Is 3 not 5.
     
  12. Dec 20, 2007 #11
    hehe i noticed a lack of parenthesis... thanks for helping :smile:
     
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