definite integrals

更新:

du/[u(u - 1)]

=[1/(u - 1) - 1/u] du?

更新 2:

but the answer is 1

3 個解答

評分
  • 最愛解答

    Sub u = 1 + ex-1, then

    du = ex-1 dx

    dx = du/ex-1 = du/(u - 1)

    When x = 0, u = 1 + 1/e, when x = 2, u = 1 + e

    Hence

    ∫(x = 0 → 2) dx/(1 + ex-1)

    = ∫(u = 1 + 1/e → 1 + e) du/[u(u - 1)]

    = ∫(u = 1 + 1/e → 1 + e) [1/(u - 1) - 1/u] du

    = [ln |u - 1| - ln |u|] (u = 1 + 1/e → 1 + e)

    = [ln |1 - 1/u|] (u = 1 + 1/e → 1 + e)

    = ln |-e| - ln |-1/(1 + 1/e)|

    = 1 - ln [e/(e + 1)]

    = 1 - ln e + ln (e + 1)

    = ln (e + 1)

    2010-12-29 22:06:05 補充:

    This is the method of partial fraction

    2010-12-30 19:45:55 補充:

    Redo from this step:

    [ln |1 - 1/u|] (u = 1 + 1/e → 1 + e)

    When u = 1 + 1/e:

    1 - 1/u = 1 - 1/(1 + 1/e)

    = 1 - e/(1 + e)

    = 1/(1 + e)

    When u = 1 + e:

    1 - 1/u = 1 - 1/(1 + e)

    = e/(1 + e)

    Therefore:

    [ln |1 - 1/u|] (u = 1 + 1/e → 1 + e)

    = ln |e/(1 + e)| - ln |1/(1 + e)|

    = ln e

    = 1

    資料來源: 原創答案
  • 1 十年前

    how do you evaluate it?

  • 計算有誤! The answer is 1.

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