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Need help (Integral)

  1. Aug 7, 2005 #1
    Hey, everyone

    I am working on a calc problem, and I have no idea where to start. The integral is

    e^x
    ------------- [division problem]
    (25+e^2x)^4

    Do I let my u equal to the 25+e^2x? or what...
    then after that what do i do.

    Thanks for all the help in advance.
    -Eiano
     
  2. jcsd
  3. Aug 7, 2005 #2

    GCT

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    here's an ingenious idea, try it out, yes try the substitution you mentioned and we'll see what magically happens.
     
  4. Aug 7, 2005 #3
    Anyone else?
    (i KNOW my u can't be 25+e^2x, that's what th OR WHAT was for)
    but HAHA, u were so funny with that response.. haha
     
  5. Aug 7, 2005 #4

    LeonhardEuler

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    Partial fraction decomposition? It looks like it'll take a long time, though.
     
  6. Aug 7, 2005 #5
    He was serious.. that's the solution.
     
  7. Aug 7, 2005 #6

    GCT

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    uhmm, actually I wasn't, he he o:)

    partial fractions...hmm....sounds interesting, we'll have to try it out, although I've never done partial fractions with e^x,e^2x as a variable.

    [tex]u=(25+e^{2x}),~du=2e^{2x}dx,~e^{x}= \sqrt{u-25}[/tex]

    next substitute u=1/t

    simplifying will give you soley a square root function in the denominator, convert this to a completed square, the rest should be easy

    I'm sure there's a more elegant solution though

    latex is ****ing up, I've actually typed the whole problem and solution out through latex, however for some reason it's picking up some old latex data from a couple of months ago :confused:
     
  8. Aug 7, 2005 #7

    GCT

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    alright finally got the latex to appear

    [tex]I = \int \frac{e^{x}dx}{(25+e^{2x})^4}[/tex]

    [tex]u=(25+e^{2x}),~du=2e^{2x}dx,~e^{x}= \sqrt{u-25}[/tex]

    [tex]I=.5 \int \frac{du}{u^{4} \sqrt{u-25}} [/tex]

    [tex]u=1/t,~du=-1/t^{2} dt[/tex]

    [tex]I=-.5 \int \frac{dt}{ \sqrt{1/t^{2} - 25/t}}[/tex]

    [tex] -.5 \int \frac{dt}{ \sqrt{(1/t -12.5)^{2} - 12.5^{2}}} [/tex]

    [tex] (1/t -12.5) = 12.5sec \theta [/tex]

    the rest should be easy

    etc...

    see any errors, please point them out, I'm guessing that there's probably a more elegant solution
     
    Last edited: Aug 7, 2005
  9. Aug 7, 2005 #8

    HallsofIvy

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    But e2x is (ex)2 so 25+ e2x= 25+ (ex)2. Try letting u= ex first, so that du= exdx. Now the integral is [tex]\int \frac{du}{25+u^2}[/tex] and that smells like an arctangent. Let u= 5 v so that du= 5 dv and 25+ u2= 25(1+ v2). Now the integral is [tex]\frac{5}{5}\int \frac{dv}{v^2+1}= arctan(v)+ C[/tex].
    Of course, that's [tex]arctan(\frac{u}{5})+ C= arctan(\frac{e^x}{5})+ C[/tex]
     
    Last edited: Aug 7, 2005
  10. Aug 7, 2005 #9

    TD

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    Unfortunately, there was a 4th power in the denominator... :wink:
     
  11. Aug 7, 2005 #10

    GCT

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    I'm not quite sure what you're getting at there can you clarify? also note as TD said, there's a fourth power in the denominator.

    I'm not able to see anything wrong with my solution.
     
  12. Aug 7, 2005 #11

    GCT

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    HallsofIvy, nevermind, I thought you were pointing out an error in my solution
     
  13. Aug 7, 2005 #12

    TD

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    If it weren't for the fourth power, HallsofIvy would've had an easy solution. He just must have missed it, I don't think he meant that yours was wrong. Without the fourth power, yours just seemed so long :smile:
     
  14. Aug 7, 2005 #13

    GCT

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    his method might lead to a partial fractions solution
     
  15. Aug 7, 2005 #14

    LeonhardEuler

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    I think I might have found an error here, GCT. Substituting gives:
    [tex]I=.5 \int \frac{(-1/t^2)dt}{ (1/t)^4\sqrt{1/t - 25}}[/tex]
    [tex]=-.5 \int \frac{dt}{ (1/t)^2\sqrt{1/t - 25}}[/tex]
    [tex]=-.5 \int \frac{dt}{ \sqrt{1/t^5 - 25/t^4}}[/tex]
     
  16. Aug 7, 2005 #15

    GCT

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    I'm not quite sure what you did there, note that bringing in 1/t^2 within the square root will reduce it to 1/t within the square root.
     
  17. Aug 7, 2005 #16

    LeonhardEuler

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    Doesn't it go like this:
    [tex]\sqrt{a}\sqrt{b}=\sqrt{ab}[/tex]
    [tex]\frac{1} {t^2}\sqrt{x} = \sqrt{\frac{1} {t^4}}\sqrt{x}=\sqrt{\frac{x} {t^4}}[/tex]
     
  18. Aug 7, 2005 #17

    GCT

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    yeah, you're right
     
  19. Aug 7, 2005 #18

    GCT

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    alright, shame on me

    What we can do is

    [tex]I=.5 \int \frac{du}{u^{4} \sqrt{u-25}} [/tex]
    [tex]u=25sec^{2} \theta ,~du=25tan \theta d \theta [/tex]
    [tex]I= \frac{-1}{10(25^{3})} \int cos^{8} \theta d \theta [/tex]
    which can be solved using standard procedure ("table integral")

    hopefull I haven't goofed up this time, I'll be posting the full version later
     
    Last edited: Aug 7, 2005
  20. Aug 7, 2005 #19

    LeonhardEuler

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    Oh, man bad news, GCT. If [itex]u=\sec^2{\theta}[/itex] then
    [tex]\frac{du}{d\theta}=\tan{\theta}\sec^2{\theta}[/tex]
    you were probably thinking of
    [tex]\int \sec^2{\theta}=\tan{\theta}[/tex]
    It's Ok, we all have bad days!
     
  21. Aug 7, 2005 #20

    GCT

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    I think I've been drinking way too much these days, anyways thanks for pointing that out.

    So the modification would result in
    [tex]I= \frac{-1}{10(25^{3})} \int cos^{6} \theta d \theta [/tex]
    ...right? :wink:
     
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