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Math Help - Integration by parts

  1. #1
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    6

    Integration by parts

    no.1 on Flickr - Photo Sharing!
    I can't get the answers they give. I keep getting

    (1) 1/2 x^2 lnx - 5x lnx - 1/4 x^2 + 5x + c

    u=lnx
    du=1/x dx

    dv=x-5
    v=x^2/2 - 5x

    lnx (x^2/2 - 5x) - int (x^2/2 -5x)(1/x) dx
    lnx (x^2/2 - 5x) - int (x/2 - 5) dx
    lnx (x^2/2 - 5x) - x^2/4 + 5x + C
    1/2x^2 lnx - 5x lnx - 1/4 x^2 + 5x +C
    What am I doing wrong?

    Also this one has been giving me trouble
    http://www.flickr.com/photos/47939227@N03/4528886766/
    I keep getting
    2/3 x (x^2+6)^3/2 - 4/5(x^2+6)^5/2 + C

    (2) int (2x^3) (x^2 + 6)^1/2
    u=2x^3
    du=6x^2 dx

    dv= (x^2 + 6)^1/2
    v=(x^2 + 6)^3/2
    --------------
    3x
    2/3 x (x^2 + 6)^3/2 - int ((x^2 + 6)^3/2 (6x^2)) /(3x)

    2/3 x (x^2 + 6)^3/2 - int (x^2 + 6)^3/2 (2x)
    2/3 x (x^2 + 6)^3/2 - 4/5 x (x^2 + 6)^5/2 + C
    What am I doing wrong with this one?

    Thank you very much for any help.
    Last edited by bananasinpajamas; April 20th 2010 at 10:07 AM.
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  2. #2
    Super Member Deadstar's Avatar
    Joined
    Oct 2007
    Posts
    722
    Quote Originally Posted by bananasinpajamas View Post
    no.1 on Flickr - Photo Sharing!
    I can't get the answers they give. I keep getting

    (1) 1/2 x^2 lnx - 5x lnx - 1/4 x^2 + 5x + c

    u=lnx
    du=1/x dx

    dv=x-5
    v=x^2/2 - 5x

    lnx (x^2/2 - 5x) - int (x^2/2 -5x)(1/x) dx
    lnx (x^2/2 - 5x) - int (x/2 - 5) dx
    lnx (x^2/2 - 5x) - x^2/4 + 5x + C
    1/2x^2 lnx - 5x lnx - 1/4 x^2 + 5x +C
    What am I doing wrong?
    Nothing. That's the answer. I think wherever you took the question from has got it wrong.

    Maple and Wolfram both give 1/2x^2 lnx - 5x lnx - 1/4 x^2 + 5x as the answer.
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  3. #3
    MHF Contributor harish21's Avatar
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    Also this one has been giving me trouble
    no.3 on Flickr - Photo Sharing!
    I keep getting
    2/3 x (x^2+6)^3/2 - 4/5(x^2+6)^5/2 + C

    (2) int (2x^3) (x^2 + 6)^1/2
    u=2x^3
    du=6x^2 dx

    dv= (x^2 + 6)^1/2
    v=(x^2 + 6)^3/2
    --------------
    3x
    2/3 x (x^2 + 6)^3/2 - int ((x^2 + 6)^3/2 (6x^2)) /(3x)

    2/3 x (x^2 + 6)^3/2 - int (x^2 + 6)^3/2 (2x)
    2/3 x (x^2 + 6)^3/2 - 4/5 x (x^2 + 6)^5/2 + C
    What am I doing wrong with this one?

    Thank you very much for any help.[/QUOTE]

    for No.2

    Avoiding integration by parts:

     2 \int {x^3} {\sqrt(x^2+6)} dx

    Let  u = x^2 \implies  du = 2 x dx :

     =  \int u {\sqrt{u+6}} du

    Again, let  v = u+6  \implies dv =  du

     =  \int (v-6) \sqrt v dv

      = \int (v^{\frac{3}{2}}-6 \sqrt v ) dv

    now integrate and then substitute the values of v and u
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