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Hard integral to me but you?

  1. Apr 22, 2005 #1
    (edited need help) hard integral to me but you?

    I tried much but i couldn't resolve.. :grumpy:
    Integral of (1+ln(x))/xln(x) dx
    thanks =)
     
    Last edited: Apr 22, 2005
  2. jcsd
  3. Apr 22, 2005 #2

    arildno

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    Hint:
    What is
    [tex]\frac{d}{dx}(xlnx)=?[/tex]
     
  4. Apr 22, 2005 #3
    its 1/x right?
     
    Last edited: Apr 22, 2005
  5. Apr 22, 2005 #4

    arildno

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    No, you must use the product rule:
    [tex]\frac{d}{dx}(xln(x))=1*ln(x)+x*\frac{1}{x}=ln(x)+1[/tex]
    What does that tell you?
     
  6. Apr 22, 2005 #5
    oh you are right!
    I know the answer then, BUT
    I WROTE MY EXERCISE BAD SORRY HERE GO AGAIN
    integral of (sqrt(1+ln(x)))/xln(x) dx
     
    Last edited: Apr 22, 2005
  7. Apr 22, 2005 #6

    GCT

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    Does this pertain to finding the arc length of a function? Also, try writing it out in latex, you'll probably get more responses.
     
  8. Apr 22, 2005 #7
    is this the question??

    [tex] \frac{\sqrt{1+\ln x}}{x \ln x} dx [/tex]?

    in which case substitute ln x = u and solve away!
     
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