Question

∫dx/√(x^2+1)^5 求不定积分的详细过程 谢谢

Answer 1

如上图所示。

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谢谢 看懂了

Answer 2

令x=tanθ
x^2+1 = (tanθ)^2+1 = (secθ)^2
√(x^2+1)^5 = (secθ)^5
dx = d(tanθ) = (secθ)^2dθ
∫dx/√(x^2+1)^5 = ∫(secθ)^2dθ/(secθ)^5 = ∫dθ/(secθ)^3 = ∫(cosθ)^3dθ
∫(cosθ)^3dθ = ∫[1-(sinθ)^2]cosθdθ = ∫[1-(sinθ)^2]dsinθ = sinθ - (sinθ)^3/3 + C
由x=tanθ，知sinθ = x/√(x^2+1)
∫dx/√(x^2+1)^5 = x/√(x^2+1) + x^3/[3(x^2+1)^3/2] + C

Answer 3

• 令x=tant

  1+x²=1+tan²t=sec²t

  dx=sec²t·dt

  ∫√(x²+1)⁻⁵dx=∫sec⁻⁵t·sec²t·dt=∫sec⁻³t·dt=∫cos³tdt

• ∫cos³tdt

  =∫(1-sin²t)·cost·dt

  =∫(1-sin²t)·dsint

  =sint-⅓sin³t+C

• 原式=sin(arctanx)-⅓sin³(arctanx)+C

  =x/√(1+x²)-⅓[x/√(1+x²)]³+C