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Integration By Parts-Is my answer correct?

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

    ∫9(ln9x)2dx

    2. Relevant equations



    3. The attempt at a solution
    u=(ln9x)2 dv=9dx
    du=2(ln9x)*1/xdx v=9x

    After all my work I get:
    9x(ln9x)2-18(ln9x)+18x

    Webassign tells me I'm wrong, but I have worked it 3 times to the same answer.
    Please help me.
    Neil
     
  2. jcsd
  3. Aug 29, 2010 #2
    Your substitution is all right. Check du, however. You have to use the chain rule twice.
     
  4. Aug 29, 2010 #3
    I did a 2nd substitution, but I was trying to save writing out the lines. Here is my work:
    9x(ln9x)2-18∫xln9x*1/xdx
    9x(ln9x)2-18∫ln9x dx (here the x * 1/x cancelled)

    u=ln9x dv=dx
    du=1/xdx v=x

    9x(ln9x)2-18[xln9x-∫dx]
    9x(ln9x)2-18[xln9x-x]

    =9x(ln9x)2-18xln9x+18x+C

    So, is this correct or where is my error?
     
  5. Aug 29, 2010 #4
    The derivative of ln 9x is not 1/x, it's 9/x by the chain rule.
     
  6. Aug 29, 2010 #5

    jgens

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    Gold Member

    Not true. The derivative is in fact 1/x if you apply the chain rule correctly.
     
  7. Aug 29, 2010 #6

    jgens

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    Gold Member

    This is the answer that I got.
     
  8. Aug 29, 2010 #7
    Cool, then it's the fact that webassign hates a Mac. Thanks
    Neil
     
  9. Aug 29, 2010 #8
    put u=9x
    du/9=dx
    so integral (lnu)2 du
    put v=ln(u)
    dv=du/u
    since u=ev
    so
    evdv=du

    integral v2 evdv
    v2ev-2vev+2ev+c
    so 9x(ln(9x))2-18xln9x+18x+c
     
  10. Aug 29, 2010 #9
    The bolded part should be 18x. Maybe you put in 18 instead of 18x so you lost points for it.
     
  11. Aug 29, 2010 #10

    jgens

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    If you choose u=log(9x) from the start, you can reduce the number of computations that you need to reach exp(u)u2.
     
  12. Aug 29, 2010 #11
    LEPTOS, in my work on paper I have 18x and I put that in webassign. I really!!!!can't type. My math is better than my typing if that says anything.
     
  13. Aug 29, 2010 #12
    The very first set of u and du are not actually substitutions (or at least as I was told by my professor). Whom ever created our textbook could find a better letter combo in the entire 26 letters of the alphabet. Go figure. The use of u and du in integration by parts has caused me pain to no end because I want to pull the 2 in the du and get 1/2du=(ln9x)*1/xdx...but apparently that is wrong.
     
  14. Aug 29, 2010 #13
    Yeah, um, wow. [sackcloth and ashes].
     
  15. Aug 30, 2010 #14

    HallsofIvy

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    Science Advisor

    You can find the derivative of f(x)= ln(9x) in either of two ways:

    chain rule: let u= 9x. Then du/dx= 9 and f(u)= ln(u) so df/du= 1/u= 1/(9x). Now df/dx= (df/du)(du/dx)= (9)(1/(9x))= 1/x.

    The easy way: f(x)= ln(9x)= ln(x)+ ln(9). Since ln(9) is a constant, df/dx= d(ln(x))/dx+ 0= 1/x.
     
  16. Aug 30, 2010 #15
    Just because I can say this, and laugh about it...HallsofIvy....the easy way....calculator. j/k
     
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