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Integration of powers of natural logs

  1. Sep 15, 2004 #1

    I need some help! :uhh:

    I have a series of maths problems in my physics work...

    I am trying to integrate a power of a log - such as:

    y= 3.2875E+004 - 8.9321E+005*ln x + 4.2316E+006*(ln x)^2 - 7.7776E+006*(ln x)^3 + 7.4949E+006*(ln x)^4 - 4.1373E+006*(ln x)^5 + 1.3164E+006*(ln x)^6 - 2.2448E+005*(ln x)^7 + 1.5873E+004*(ln x)^8

    Any help at all would be great :smile:

    All I have is that: integral of lnx = xlnx - x + C

    even just the simple y = 2*(lnx)^3, would be great, I can do the rest from there...I assume its a 'by parts' expression?

    Cheers :redface:
  2. jcsd
  3. Sep 15, 2004 #2


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    Sounds reasonable; have you tried it?
  4. Sep 16, 2004 #3
    Well I was thinking - but I don't know if its right and I don't know where to find out - hence posting here... :smile:

    y = 2*(lnx)^3

    = 2*[(lnx)(lnx)(lnx)]

    therefore integral = 2*( xlnx - x * xlnx - x * xlnx - x) +C
    = 2*(3xlnx - x) + c
    =6xlnx - x + c

    Is that right?

    Thanks very much for the help. :biggrin:
  5. Sep 16, 2004 #4


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    This is absolutely wrong!!
    (Ask yourself: How can I prove that my answer is wrong? Hint: Differentiate..)
  6. Sep 16, 2004 #5
    ok, so can you help me get further than this? maybe pointing me in the direction of the parts bit?

    I know that: (all this is from the web, I have done this work - but 6 years ago..)

    integral = $

    $udv = uv - $vdu

    [eg $xe^x.dx, u = x, dv = e^x.dx]

    so for the expression: $2(lnx)^3.dx

    I need a u part and a dv part:

    u = lnx, (dv = x^3.dx)??? I'm not sure what the dv part should be??

    I also assume I take the constant out?

    Cheers :redface:
  7. Sep 16, 2004 #6


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    Set u=(ln(x))^3 v'=1
    uv=xln(x)^3, vu'=3ln(x)^2
    Then, use the same technique with (ln(x))^2
  8. Sep 16, 2004 #7
    Why is dv/dx = 1?

    so here is my attempt at an answer:


    u=(lnx)^3, v'=1

    uv = x(lnx)^3, vu'=3(lnx)^2

    u = (lnx)^2, v'=1

    uv= x(lnx)^2, vu'=2lnx

    so: $udv=x(lnx)^3 -$3(lnx)^2
    = x(lnx)^3 - x(lnx)^2 - $2lnx
    =x(lnx)^3 - x(lnx)^2 - 2(xlnx - x)

    so complete answer would be:
    = 2(x(lnx)^3 - x(lnx)^2 - 2(xlnx - x)) + C ??
  9. Sep 16, 2004 #8
    I really am confused with the v' =1 i.e. are you saying that the integral of lnx = 1?? :cry:

    How can that be - as when you differentiate you do not get lnx???

    If you are not saying that then what is going on = can you break it down for me please as I'm very confused...

    I thought the integral of lnx = xlnx - x +C ????

    also - if you have: $2(lnx)^3 how is taking u = (lnx)^3 breaking it down - isn't that the same as what you start with - ie needing to break into parts...

    I'm sorry if this is really obvious - but I am having real trouble working this out...

  10. Sep 16, 2004 #9


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    You are mixing up dv and v'!
    The integration by parts formula reads:
    You may write this (using the change of variables formula) as:
  11. Oct 21, 2006 #10
    Cheers guys

    dats helpd a lil man... i was kinda stuck on da integral of (lnx)^2, an i thought dat mayb cos yu got (lnx) to the power 2, dat yu cnt take u = (lnx)^2... but now im guessin yu can... RIIGHT? :uhh: :confused:
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