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Integral with e

  1. Jun 10, 2009 #1
    1. The problem statement, all variables and given/known data

    [tex]\int[/tex][tex]\frac{e^{z}}{1+e^{2z}}[/tex]

    This is clearly a u substitution problem. My instinct is to let u = [tex]{1+e^{2z}[/tex]
    so du = 2e^[tex]{2z}[/tex]dz :confused:

    Now what? I don't see how this can be substituted into the original integral.

    I could let u = e^z, but then that leaves e^2z. Clearly du in this case be e^zdz, but that also doesn't substitute in. e^2z = e^z * e^z. Could this be part of the solution? If somebody could provide some hint to get me going, I'd be grateful.
     
  2. jcsd
  3. Jun 10, 2009 #2

    Dick

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    I like u=e^z better. That turns e^(2z) into u^2. Yes, that is part of the solution.
     
  4. Jun 10, 2009 #3
    I agree. Try u=exp(z). du=exp(z) dz. Then you should get a very familiar integral. (The solution will be a trig function)
     
  5. Jun 10, 2009 #4
    Okay, so if I use e^z = u and du = e^zdz, I am left with

    u/(1+u^2)dz

    This is for my calc 2 class. I have no idea how a trig function would tie into this though :(
     
  6. Jun 10, 2009 #5

    rock.freak667

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    [tex]\int \frac{e^z}{1+e^{2z}}dz[/tex]

    so yes, with u=ez => du=ez dz

    so you will be left with

    [tex]\int \frac{1}{1+u^2} du[/tex]

    NOT

    [tex]\int \frac{u}{1+u^2}du[/tex]

    can you integrate 1/(1+u2) du?
     
  7. Jun 10, 2009 #6
    Ah, you're right.

    I took calculus 1 at my college, but I am taking calculus 2 at a different school. If I understand what an integral is, then we are seeking a function whose derivative IS 1/1+x^2. My calc 1 class never covered this, but I am beggining to suspect the calc 1 at this new college did: inverse trig functions. A peek in the back of my textbook indicates that 1/1+x^2 is the derivative of the arctan function. Is that it?

    So I'm simply looking at arctan (e^z)?
     
  8. Jun 10, 2009 #7
    I appreciate all of your help, by the way. I hope I don't come across as helpless or not wanting to do the work. I've gotten through most of this assignment but may have encountered something I haven't seen before.
     
  9. Jun 10, 2009 #8

    rock.freak667

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    well yes...but arctan(ez)+C (never forget the constant of integration)
     
  10. Jun 10, 2009 #9
    Costs me 1 point every time :grumpy:
     
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