第2个回答 2020-07-27
∫(3,+∞) dx/[(x-1)^4 ∨(x²-2x)]
=∫(3,+∞) dx/[(x-1)^4 ∨((x-1)²-1)] ①
令x-1=secθ,则x=1+secθ
dx=secθtanθdθ
x∈[3,+∞),则x-1 ∈[2,+∞),即secθ∈[2,+∞)
secθ=1/cosθ
secθ=2时,cosθ=1/2,θ=π/3
secθ=+∞时,cosθ=0,θ=π/2
所以θ∈[π/3,π/2]
①=∫(π/3,π/2) secθ·tanθdθ /[(secθ)^4 ·tanθ]
=∫(π/3,π/2) cos³θdθ
=∫(π/3,π/2) cos²θdsinθ
=∫(π/3,π/2) (1-sin²θ)dsinθ
=(sinθ-1/3 sin³θ)|(π/3,π/2)
=1-1/3-∨3/2+1/3 (∨3/2)³
=2/3-3∨3/8本回答被提问者采纳