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Nightmare integral

  1. Feb 16, 2006 #1
    Hi all,
    I need ur help with this integral
    [tex]\int\frac{x*ln(x+sqrt{1+x^2}*dx} {sqrt{1+x^2} [/tex]
    maybe its about substitution but how?
     
    Last edited: Feb 16, 2006
  2. jcsd
  3. Feb 16, 2006 #2

    arildno

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    Did you mean that?
     
    Last edited: Feb 16, 2006
  4. Feb 16, 2006 #3
    exactly :)
     
  5. Feb 16, 2006 #4

    arildno

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    At first glance, integration by parts seems best, with [tex]\frac{dv}{dx}=\frac{x}{\sqrt{1+x^{2}}}[/tex]
     
  6. Feb 16, 2006 #5

    Hurkyl

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    It seems to me that there are two obvious substitutions to try. Have you tried anything yet?

    Not to mention IBP -- you've surely seen how to do the integrals of x ln x and ln x by now, and seen the theme there?
     
    Last edited: Feb 16, 2006
  7. Feb 16, 2006 #6
    by doing this I got

    [tex]\int{udv}=ln(x+t)t-\int\frac{(2t+1)xdx} {x+t}[/tex]
    where [tex]t=\sqrt{1+x^2}[/tex]

    is it right? and how then?
     
    Last edited: Feb 16, 2006
  8. Feb 16, 2006 #7

    arildno

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    Incorrect!
    We have:
    [tex]\frac{dt}{dx}=\frac{x}{t}[/tex]
    Thus, your last integral is:
    [tex]\int\frac{t(1+\frac{x}{t})}{x+t}dx=\int{dx}[/tex]
     
  9. Feb 16, 2006 #8
    I am a self-study and new to calculus and have no experience, so I hope u explain more about ur point
     
  10. Feb 16, 2006 #9

    Hurkyl

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    One of them relies on trigonometric substitution -- I suppose if you haven't gotten that far, then the relevant substitution would not be obvious. :smile:


    The other one stems from the fact you (presumably) know how to integrate things that look like "ln x", but not "ln [something other than a plain ordinary x]". So, one obvious thing to try is to do something that will turn "ln [something other than a plain ordinary x]" into something akin to "ln x".
     
    Last edited: Feb 16, 2006
  11. Feb 16, 2006 #10
    yes. there was an error in deffirentiating ln.
    I appiciate ur help arildno :)
    anyway what about other interesting methods?

    thanx alot
     
  12. Feb 16, 2006 #11

    arildno

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    Well, you might try the substitution x=Sinh(u), where Sinh() is the hyperbolic sine function.
     
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