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Integral from Hell

  1. Feb 26, 2009 #1
    Integral from Hell!!

    1. The problem statement, all variables and given/known data

    [tex]\int
    \frac{\sqrt{ln(9-x}}{\sqrt{ln(9-x} + \sqrt{ln(x+3)}}
    dx
    [/tex]


    2. Relevant equations
    woe = me


    3. The attempt at a solution
    Tried all the algerbraic manipulations I could think of, can't figure out a substitution either.
     
  2. jcsd
  3. Feb 26, 2009 #2
    Re: Integral from Hell!!

    Have you tried multipying by the conjugate? Just a guess ( not a hint guess, but a real guess)
     
  4. Feb 26, 2009 #3
    Re: Integral from Hell!!

    Tried didn't get me anywhere seemingly
     
  5. Feb 26, 2009 #4

    gabbagabbahey

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    Re: Integral from Hell!!

    Is this supposed to be an indefinite integral, or are you given limits? Also, are you sure it is exactly as written above?
     
  6. Feb 26, 2009 #5
    Re: Integral from Hell!!

    limits from 2 to 4, figured i'd just get help getting the anti derivative
     
  7. Feb 26, 2009 #6

    gabbagabbahey

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    Re: Integral from Hell!!

    I don't think there is a general antiderivative; but I suspect that the limits allow you to use some trick (like in the 1/(1+tan^b (x) ) thread).....hmmmm, let me see....
     
  8. Feb 26, 2009 #7
    Re: Integral from Hell!!

    Holy crap this is linear! anyone care to help me figure out why?
     
  9. Feb 26, 2009 #8

    gabbagabbahey

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    Re: Integral from Hell!!

    Really?!...Doesn't look exactly linear to me....It looks close to a straight line, but not really.
     
  10. Feb 26, 2009 #9
    Re: Integral from Hell!!

    I could have been mistaken
     
  11. Feb 26, 2009 #10

    gabbagabbahey

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    Re: Integral from Hell!!

    I'd start by dividing the numerator and denominator by [itex]\sqrt{\ln(9-x)}[/itex]...
     
  12. Feb 26, 2009 #11
    Re: Integral from Hell!!

    nice one! how didn't i see that!!
     
  13. Feb 26, 2009 #12

    gabbagabbahey

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    Re: Integral from Hell!!

    Lol , actually, I'm still trying to see why the area under [tex]\frac{1}{1+\left(\frac{\ln(3+x)}{\ln(9-x)}\right)^b}[/tex] is the same as the area under [tex]\frac{1}{1+\left(\frac{\ln(9-x)}{\ln(3+x)}\right)^b}[/tex]

    Is there some proof I'm not remembering?
     
  14. Feb 26, 2009 #13
    Re: Integral from Hell!!

    not that i know of, i'm kind of stuck at that point now I don't see any substitutions that would make things simpler from here.
     
  15. Feb 26, 2009 #14
    Re: Integral from Hell!!

    [tex]\frac{\frac{\ln(3+x)}{\ln(9-x)}^b}{1+\left(\frac{\ln(3+x)}{\ln(9-x)}\right)^b}[/tex]
     
  16. Feb 26, 2009 #15
    Re: Integral from Hell!!

    I dont see that going anywhere, anyone got ideas?
     
  17. Feb 26, 2009 #16

    gabbagabbahey

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    Re: Integral from Hell!!

    Sure, [tex]\frac{1}{1+\left(\frac{\ln(9-x)}{\ln(3+x)}\right)^b}= \frac{\frac{\ln(3+x)}{\ln(9-x)}^b}{1+\left(\frac{\ln(3+x)}{\ln(9-x)}\right)^b}[/tex]

    so [tex]\frac{1}{1+\left(\frac{\ln(3+x)}{\ln(9-x)}\right)^b}+\frac{1}{1+\left(\frac{\ln(9-x)}{\ln(3+x)}\right)^b}=1[/tex]

    But that doesn't explain why the area under [tex]\frac{1}{1+\left(\frac{\ln(3+x)}{\ln(9-x)}\right)^b}[/tex] is equal to the area under [tex]\frac{1}{1+\left(\frac{\ln(9-x)}{\ln(3+x)}\right)^b}[/tex] over that interval (although I know from mathematica that it is)
     
  18. Feb 26, 2009 #17

    gabbagabbahey

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    Re: Integral from Hell!!

    Hmmm....what do you get when you flip the function [tex]\frac{1}{1+\left(\frac{\ln(3+x)}{\ln(9-x)}\right)^b}[/tex] over the line [itex]x=3[/itex]?

    You may need to look in your linear algebra text, under the transformations/reflections section for this one)
     
  19. Feb 26, 2009 #18
    Re: Integral from Hell!!

    I think I'm giving up!! This was one of 4 challenge problems my professor gave... I think I got 2 of the others I don't know on this one haha.
     
  20. Feb 26, 2009 #19

    gabbagabbahey

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    Last edited by a moderator: May 4, 2017
  21. Feb 26, 2009 #20
    Re: Integral from Hell!!

    Of course not, nice job finding that out, not sure if it is gonna help me compute the integral.
     
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