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定积分∫(-1→1)1/x^2dx=

      发布时间: 2019-09-17

      π/2) sinycosy dy
      =(1/2
      ∫(1->2)∫(arctan1->π/(1+x^2)^2dx
      =∫(arctan1->+∞)x/π/4 + (1/2) sin2y dy
      =(-1/2) tany/4) [cos2y] (arctan1->(secy)^2 dy
      =∫(arctan1->2)
      =1/π/let
      x=tany
      dx=(secy)^2 dy
      x=1, y= π/, y=arctan1
      x=+∞

      回复:

      let x=tany dx=(secy)^2 dy x=1, y=arctan1 x=+∞, y= π/2 ∫(1->+∞)x/(1+x^2)^2dx =∫(arctan1->π/2) tany/(secy)^2 dy =∫(arctan1->π/2) sinycosy dy =(1/2)∫(arctan1->π/2) sin2y dy =(-1/4) [cos2y] (arctan1->π/2) =1/4 + (1/4)cos(2(arctan1))

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