第1个回答 2020-05-15
t = x/(1+x)
x = t/(1-t)
dx = 1/(1-t)^2 dt
I = ∫ [0, 3/4]√t /(1-t)^2 dt
= ∫ [0,√3/2] (2u^2-2 + 2) /(1-u^2)^2 du, where u^2 = t
= ∫ [0,√3/2] - [1/(1-u) + 1/(1+ u)] + (1/2)[1/(1+u) + 1/(1+u)^2 + 1/(1-u) - 1/(1-u)^2] du
= ∫ [0,√3/2] -(1/2) [1/(1-u) + 1/(1+ u)] + (1/2)[ 1/(1+u)^2 - 1/(1-u)^2] du
= (1/2)ln|(1-u)/(1+u)| - (1/2)[1/(1+u) + 1/(1-u)] [0,√3/2]
= Answer (2.147)本回答被网友采纳
x = t/(1-t)
dx = 1/(1-t)^2 dt
I = ∫ [0, 3/4]√t /(1-t)^2 dt
= ∫ [0,√3/2] (2u^2-2 + 2) /(1-u^2)^2 du, where u^2 = t
= ∫ [0,√3/2] - [1/(1-u) + 1/(1+ u)] + (1/2)[1/(1+u) + 1/(1+u)^2 + 1/(1-u) - 1/(1-u)^2] du
= ∫ [0,√3/2] -(1/2) [1/(1-u) + 1/(1+ u)] + (1/2)[ 1/(1+u)^2 - 1/(1-u)^2] du
= (1/2)ln|(1-u)/(1+u)| - (1/2)[1/(1+u) + 1/(1-u)] [0,√3/2]
= Answer (2.147)本回答被网友采纳
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