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Integration Problem

  1. Aug 31, 2008 #1
    1. The problem statement, all variables and given/known data

    [tex]\int\frac{1}{xln(x)^2}[/tex]

    2. Relevant equations

    [tex]\frac{d}{dx}[/tex]ln(x) = [tex]\frac{1}{x}[/tex]

    3. The attempt at a solution

    I made a u substitution, letting u = ln(x); du = [tex]\frac{1}{x}[/tex]dx. The antiderivative I then get is ln(u[tex]^{2}[/tex]) + C but obviously this is not the correct answer. Any help anyone could provide would be greatly appreciated.

    Derek
     
  2. jcsd
  3. Aug 31, 2008 #2
    Have you tried integrating by parts?
     
  4. Aug 31, 2008 #3
    Thanks for your reply. Unfortunately we haven't covered integration by parts yet; we have only made it as far as logarithmic differentiation so the techniques that I know are limited. derek
     
  5. Aug 31, 2008 #4
    the procedure you are doing seems correct but your final answer is not correct. You are therefore making an error in the process. Please post your full solution and perhaps someone will point out the error. Or if this helps the answer should be

    -1/lnx + c
     
  6. Aug 31, 2008 #5
    Integration by parts is not necessary at all. Just rework your problem with the substitution you chose, and you'll see the error you are making.
     
  7. Aug 31, 2008 #6
    Thanks for your replies guys. I'm still not seeing it. After the substitution I get [tex]\int\frac{1}{u^{2}}du = ln(u^{2}) + C = 2ln(u) + C[/tex]. I'm stumped at this point.
     
  8. Aug 31, 2008 #7
    [itex] \int x^{-2} dx = -1/x [/itex].
     
  9. Aug 31, 2008 #8
    Beautiful. I got so caught up in [tex]\frac{d}{dx}ln(x) = \frac{1}{x}[/tex] that the square didn't even register. Thanks everyone for your help. Derek
     
  10. Sep 1, 2008 #9
    answer comes out to be 1/2(ln(ln(x^2))+c
     
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