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Integration, help me please.

  1. Aug 11, 2012 #1
    1. The problem statement, all variables and given/known data

    Find:

    2. Relevant equations

    [tex]\int \frac{1}{\sqrt{e^{2x}-1}} dx[/tex]

    3. The attempt at a solution

    I tried u=e^x
     
  2. jcsd
  3. Aug 11, 2012 #2

    micromass

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    And what did you get?? Please provide a complete attempt.
     
  4. Aug 11, 2012 #3
    I think I got it :)

    [tex]\int \frac{dx}{\sqrt{e^{2x}-1}} \\ $Let u^2 = e^{2x}-1 $ \\ \therefore 2u du=2e^{2x}dx \\ udu = e^{2x}dx \\ $Now $u^2+1=e^{2x} $ from the substitution so$\\ I=\int \frac{udu}{1+u^2} \cdot \frac{1}{u} \\ = \int \frac{du}{1+u^2} \\= \tan^{-1}(\sqrt{e^{2x}-1})+C[/tex]

    Micromass, can you give me some high school level or 1st year uni integrals to practice, I did all the ones in my exercise book, I need some challenging integrals, please :)
     
  5. Aug 11, 2012 #4

    micromass

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    Looks good!

    Sure! I'll look for some challenging ones!
    If you want to get a good response, you can always make a thread in https://www.physicsforums.com/forumdisplay.php?f=109 asking for some challenging integrals!

    Let me look for some goodies.
     
  6. Aug 11, 2012 #5
    Try this one:
    [tex]
    \int{\frac{dx}{\sqrt{1 + x^4}}}
    [/tex]
     
  7. Aug 11, 2012 #6

    micromass

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    Here are some nice ones:

    [tex]\int \sqrt{\tan(x)}dx[/tex]

    [tex]\int e^{\sin(x)}\frac{x\cos^3(x)-\sin(x)}{\cos^2(x)}dx[/tex]

    [tex]\int \frac{1}{1+\sin(x)}dx[/tex]

    [tex]\int \frac{5x^4+1}{(x^5+x^+1)^2}dx[/tex]
     
  8. Aug 11, 2012 #7
    Thank you a lot, I will try my best, and then post my solutions here so you guys can check if they are right or wrong.
     
  9. Aug 11, 2012 #8
    (1)Answer for the first 1: [tex]\frac{log(sex(x)(sin(x)+cos(x))(\sqrt{2}\sqrt{tan(x)}+1)))}{2\sqrt{2}}+C[/tex] The working was very tedious, very hard integral.
    (2) Hard :(
    (3) Using a Weierstrass substitution, the answer is: [tex]-\frac{1}{t+1}+C[/tex]
    (4) Simple U-Substitution, let u=x^5+x^+1

    I couldn't show full working because I'm very tired now :(
     
  10. Aug 11, 2012 #9
    Tried a lot with this one, but it is impossible !!!
    Please show me a short simple way of doing it.
     
  11. Aug 11, 2012 #10

    micromass

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    I don't think that his integral even has an elementary solution... Maybe he made a typo??

    What about this one:

    [tex]\int \frac{1-4x^5}{(x^5-x+1)^2}dx[/tex]

    This has a very easy integral and it's obvious once you see it. But it's pretty hard to find.
     
  12. Aug 13, 2012 #11
    So how do you do this integral? It looks easy but substitutions aren't working :$
     
  13. Aug 13, 2012 #12
    What substitutions did you try?
     
  14. Aug 13, 2012 #13

    micromass

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    Substitutions won't help you. Think about the quotient rules for derivatives.
     
  15. Aug 13, 2012 #14
    micromass, I can see the solution because it fits the pattern of the quotient rule but I do not see any way of deriving it using normal integration rules. Partial fractions does not work, substitution is a dead end, integration by parts requires you to integrate an impossible function (or its just me who is not cunning enough to see how to integrate by parts correctly.) Are you aware of any such method?
     
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