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原函数求错了！应为：
令 √[1-e^(-2x)] = u, 则 e^(-2x) = 1-u^2, x = -(1/2)ln(1-u^2),
dx = .udu/(1-u^2)
I = ∫<0, √3/2> u^2du/(1-u^2) = ∫<0, √3/2> (u^2-1+1)du/(1-u^2)
= ∫<0, √3/2> [-1+1/(1-u^2)]du
= -√3/2 + (1/2) ∫<0, √3/2> [1/(1-u)+1/(1+u)]du
= -√3/2 + (1/2)[ln{(1+u)/(1-u)}]<0, √3/2>
= -√3/2 + (1/2)ln[(1+√3/2)/(1-√3/2)]
= ln[(2+√3) - √3/2
令 √[1-e^(-2x)] = u, 则 e^(-2x) = 1-u^2, x = -(1/2)ln(1-u^2),
dx = .udu/(1-u^2)
I = ∫<0, √3/2> u^2du/(1-u^2) = ∫<0, √3/2> (u^2-1+1)du/(1-u^2)
= ∫<0, √3/2> [-1+1/(1-u^2)]du
= -√3/2 + (1/2) ∫<0, √3/2> [1/(1-u)+1/(1+u)]du
= -√3/2 + (1/2)[ln{(1+u)/(1-u)}]<0, √3/2>
= -√3/2 + (1/2)ln[(1+√3/2)/(1-√3/2)]
= ln[(2+√3) - √3/2
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