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Integration by parts trial and error

  1. Sep 1, 2008 #1
    1. The problem statement, all variables and given/known data
    [tex]\int[/tex]ln(7x+9)dx

    2. Relevant equations
    derivative of ln is 1/x


    3. The attempt at a solution
    Well im just learning IBP so i set u=7x+9 dv=lndx but im stuck there. How do you know which to make ur u is there a way or is it trial and error
    Can i split the integral as:

    ∫ln(7x)+∫9
     
    Last edited: Sep 1, 2008
  2. jcsd
  3. Sep 1, 2008 #2

    Defennder

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    If you're doing by parts, note that the integrand expression (ln(7x+9)) must be comprised of udv, where u*dv. So u cannot be inside the ln expression. Let u=ln(7x+9). Continue from here.

    In general, you should let u be an expression easy to differentiate and dv be the expression easy to integrate, and make sure that vdu is easy to integrate as well. It's some trial and error which gets better with experience.
     
  4. Sep 1, 2008 #3
    u=ln(7x+9)
    du=1/(x+9)dx

    dv=dx
    v=x

    using the IBP formula i get:

    ln(7x+9)x- (1/7)∫x*(1/x+9) dx

    is this right or have i made a mistake
     
  5. Sep 1, 2008 #4

    Defennder

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    du is wrong.
     
  6. Sep 1, 2008 #5
    du=1/(x+9)*7 dx

    ok so my answer i got is

    1/7(7x+9)(log(7x+9)1)
     
  7. Sep 1, 2008 #6

    Defennder

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    Your du is still incorrect. How do you differentiate ln(f(x))?
     
  8. Sep 1, 2008 #7
    what do u mean how should it be
     
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