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Integration by parts evaluation

  1. Jan 16, 2012 #1
    ∫xax

    u=x
    du=dx
    dv=axdx
    v=ax/lna

    = xax - ∫axdx/lna

    is my solution right?
    my problem now is how to integrate the expression xax - ∫axdx/lna
    please help..
     
  2. jcsd
  3. Jan 16, 2012 #2
    Seems correct to me, you just need to evaluate the 2nd term, the integral, you know how to do integrate a^x as you've already done it, 1/ln(a) is just a constant.
     
  4. Jan 16, 2012 #3
    ok this what i've got...

    =xax - ∫axdx/lna

    =xax - 1/lna∫axdx

    =xax - (1/lna) (ax/lna)

    =xax - ax/2lna

    is that right that (lna)(lna) = 2lna?
    or it is just (lna)(lna) = (lna)(lna)
     
  5. Jan 16, 2012 #4
    It's the bottom line as:

    2ln(a) = ln(a^2)

    What you have is ln(a)ln(a) = (ln(a))^2.
     
  6. Jan 16, 2012 #5
    but , was my answer correct?
     
  7. Jan 16, 2012 #6
    xa^x - a^x/(2lna) isn't correct, xa^x - a^x/(ln(a))^2 is.
     
  8. Jan 16, 2012 #7
    ah ok..thank you very much!
     
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