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Integration x^-1

  1. Apr 19, 2005 #1
    find
    integration x^-1 :bugeye: :bugeye: :bugeye:
    i tried but reached nothing .....
    plz explain
     
  2. jcsd
  3. Apr 19, 2005 #2
    It's one of the ways to define ln(x). (Well, you use a definite integral for that, but it's practically the same).

    [tex]\int_1^x{\frac{1}{t}dt}=\ln \left|x\right|[/tex]
     
  4. Apr 19, 2005 #3

    dextercioby

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    You have to be kidding,right...?

    Are u talking about this

    [tex] \int \frac{dx}{x} [/tex]

    ,or is it a joke...?

    And what's this doing in the "Brain Teaser forum"...?April 1-st was a while ago...:rolleyes:

    Daniel.
     
  5. Apr 19, 2005 #4
    Dex you can be really mean sometimes.
     
  6. Apr 19, 2005 #5

    dextercioby

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    I have a mean looking avatar,don't I ? Why should i be gentle then ?

    I appologize to the OP,whose name i don't know and couldn't possibly find out...


    Daniel.
     
  7. Apr 19, 2005 #6
    Hehe, is that piccolo?
     
  8. Apr 19, 2005 #7

    dextercioby

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    No,he's a waiter...

    Can the OP compute the derivative of [itex] \ln x [/itex] ...?It would be useful to compute the antiderivative...

    Daniel.
     
  9. Apr 19, 2005 #8

    Galileo

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    No man, can't you see? That's Kami! Ofcourse he merged with Piccolo so now they are actually one. Good old Kami-sama was always pretty gentle actually. It was Piccolo who was mean.
     
  10. Apr 19, 2005 #9
    could u prove it plz ?
     
  11. Apr 19, 2005 #10

    dextercioby

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    It can be shown starting with the definition that

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

    Then

    [tex] \int \frac{dx}{x} [/tex]

    becomes after the substitution

    [tex] \frac{1}{x}=e^{t} \Rightarrow dx=-e^{-t} dt [/tex]

    [tex] \int e^{t}\left(-e^{-t} \ dt\right) =-\int dt=-t+C [/tex]

    Inverting the substitution,one finds

    [tex] \int \frac{dx}{x}=-\ln\left(\frac{1}{\left|x\right|}\right) +C =\ln\left|x\right| +C [/tex]


    Daniel.
     
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