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Really hard integrals?

  1. Jun 11, 2007 #1
    can someone give me some really hard intergrals to solve?

    make sure they are in the range of calculus 1-2 (anything before multivariable)

    My teacher assigned some few hard integrals, and they are fun. I want to try moer.
  2. jcsd
  3. Jun 11, 2007 #2


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    Try [tex] \int{\frac{(1+x^{2})dx}{(1-x^{2})\sqrt{1+x^{4}}}} [/tex]
    (forgot to put the integral sign in, it is now fixed)
    Last edited: Jun 11, 2007
  4. Jun 11, 2007 #3
  5. Jun 11, 2007 #4
    [tex]\int e^{-x^2} dx[/tex]
  6. Jun 11, 2007 #5


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    I doubt that it belongs to either Calculus 1 or Calculus 2 problems. :bugeye:

  7. Jun 11, 2007 #6

    Gib Z

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    [tex]\int^{1}_0 \frac{\log_e (1+x)}{x} dx [/tex]. Quite an interesting one that someone gave to me. Nice Solution :)
  8. Jun 11, 2007 #7


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    Find [tex]\frac{f'(x)}{f(x)}[/tex] where [tex] f(x) = sec(x)+tan(x) [/tex]and hence find [tex] \int sec(x) dx[/tex].

    One of my faves :smile:
  9. Jun 11, 2007 #8
    wow, these loko pretty fun. THey look way different from the ones ive ever seen.

    Ill give them a whirl sometime soon.
  10. Jun 11, 2007 #9
    This is a pretty hard one but I haven't finished Calc 2 so I don't know any harder than this.

    My favorite Integral so far is this:

    [tex]\int \frac{dx}{(x^2+9)^3}[/tex]

    It's general form is of
    [tex]\int \frac{dx}{(x^2+a^2)^n}[/tex]

    It has a really interesting answer
  11. Jun 12, 2007 #10
    Hard ,but famous and bautiful :

  12. Jun 13, 2007 #11
    Try this one...:biggrin:

    [tex]\int \sqrt{\tan(x)}{\rm dx}[/tex]
  13. Jun 13, 2007 #12
    That's a good one :rofl:
  14. Jun 14, 2007 #13
    Took me 5 minutes only :rolleyes:
    That question though, however, was just..simply amazing.
    I suggest everyone try that question
  15. Jun 14, 2007 #14

    Gib Z

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    I think the original poster has quite enough thanks...he hasn't actually done any of them yet.
  16. Jun 14, 2007 #15
    I'm stumped but intrigued.
  17. Jun 14, 2007 #16
    [tex]\int \frac{1}{x^5+1}dx[/tex]
  18. Jun 14, 2007 #17
    We are not all as clever as you Invictious :tongue:
  19. Jun 14, 2007 #18
    I'm just out of Calc 1, so I'm not sure if I even have the knowledge to solve this... but here's where I am now. I can't tell if I complicated it even more, or if I'm closer to getting the solution.

    [tex]\int (r-1)^{1/5}ln(r) - r^{-1} (r-1)^{1/5} dr[/tex]

    I used u-substitution (well, r-substitution), where [tex]r = x^5 + 1[/tex]. After the substitution I used integration by parts, and now I'm unsure if that was even the right path. Please let me know!

    And please tell us how to do [tex]\int_{0}^{\infty}sin(x^2)dx[/tex]
    Last edited: Jun 14, 2007
  20. Jun 14, 2007 #19
    how can this even be integrated?:uhh:
  21. Jun 14, 2007 #20
    It can be proved that there's no elementary antiderivative, but you can use a trick from multivariable calculus involving a change to polar coordinates and the squeeze theorem to evaluate it. It's called a Gaussian integral.

    Edit: Correction--the trick works for [tex]\int_{- \infty}^{\infty} e^{-x^{2}}dx[/tex]
    Last edited: Jun 14, 2007
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