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Three integrals to evaluate

  1. Sep 4, 2009 #1
    1. The problem statement, all variables and given/known data

    1. [tex]\int\frac{2}{e^{-x} + 1}[/tex]

    [tex]\int\frac{2t - 1}{t^{2} + 4}[/tex]

    [tex]\int\frac{4}{4x^{2} + 4x + 65}[/tex]

    2. Relevant equations



    3. The attempt at a solution

    1. I'm not sure what to do. A u-substitution would give me e^(-x) dx, but I don't see how I could get an e^(-x) dx anywhere in the integrand.

    2. Again I have no idea. I'm thinking maybe it will involve arc tan, except the numerator isn't 1, so what would I do with the 2t-1 numerator?

    3. I tried completing the square, to get something that resembled an arc tan, but that didn't work too well. I'm not sure what else to do; a u-substitution wouldn't work I don't think that would just get 8x + 4.

    No integration by parts or partial fractions allowed yet. Any help appreciated.
     
  2. jcsd
  3. Sep 4, 2009 #2

    zcd

    User Avatar

    For the first one, note that [tex]\int\frac{2}{e^{-x} + 1}\,dx=\int \frac{2}{1+\frac{1}{e^x}}}\,dx[/tex].

    For the second one [tex]\int\frac{2t - 1}{t^{2} + 4}\,dt = \int\frac{2t}{t^{2} + 4}\,dt - \int\frac{1}{t^{2} + 4}\,dt[/tex]

    For the third one, you almost have it. Try doing 2x+1 = 8tanθ
     
  4. Sep 4, 2009 #3
    Thanks for the other two hints, I don't understand what you mean by 2x+1 = tan theta?
     
  5. Sep 4, 2009 #4

    zcd

    User Avatar

    Are you familiar with trigonometric substitution?
    [tex]\int\frac{4}{4x^{2} + 4x + 65}\,dx=\int\frac{4}{(2x+1)^{2}+8^{2}}\,dx=\frac{1}{16}\int\frac{1}{\tan^{2}(\theta)+1}\,dx[/tex]

    because
    [tex]8\tan(\theta)=2x+1[/tex], [tex]\theta=\arctan(\frac{2x+1}{8})[/tex] and [tex]2dx=8\sec^{2}(\theta)[/tex].
     
  6. Sep 4, 2009 #5
    Oooh nevermind I got it
     
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