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How do I Integrate this! u substitution with limits.

  1. Feb 21, 2012 #1
    1. The problem statement, all variables and given/known data

    Find ∫e^x/ (1+e^2x). dx , with limits ln 2 & 0
    given u= e^x

    2. Relevant equations



    3. The attempt at a solution

    u= e^x
    du/dx = e^x
    dx= du/e^x

    sub limits of ln2 & 0 → u

    Hence, limits 2 & 1

    Therefore,

    ∫u* (1+e^2x)^-1* du/e^x

    = ∫ u/ (u + e^3x)
    = ∫ u/ e^3x
    = ∫ 1/e^2x

    = -e^-x
    = -1/u
    plugging in limits of 2 &1

    Therefore, 0.2325...

    Although i could not find this on the answer sheet did i do something wrong?
    Please help, Thankyou.
     
  2. jcsd
  3. Feb 21, 2012 #2
    Should it become this?
    [itex]\int^{2}_{1}(1+u^{2})^{-1}du[/itex]
    = [itex]\left[\frac{ln(1+u^{2})}{2u}\right]^{2}_{1}[/itex]
     
  4. Feb 21, 2012 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Yes, this is correct. Letting [itex]u= e^x[/itex], [itex]du= e^xdx[/itex] so the numerator is just du and [itex](1+ e^{2x}[/itex] becomes [itex]1+ u^2[/itex]

    However, this is incorrect. Yes, [itex]\int 1/u du= ln|u|+ C[/itex] but if you have a f(u) rather than u, you cannot just divide by f'(u)- that has to be already in the integral in order to make that substitution.

    Instead, look up the derivative of arctan(u).
     
  5. Feb 21, 2012 #4
    Oops haha
    should let u = tanθ
     
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