令 √(e^x-1) = u, 则 e^x = 1+u^2,
x = ln(1+u^2), dx = 2udu/(1+u^2)
I = ∫<0, 2> 2u^2du/(1+u^2)
= 2∫<0, 2> [1 - 1/(1+u^2)] du
= 2[u - arctanu]<0, 2> = 2(2-arctan2)
x = ln(1+u^2), dx = 2udu/(1+u^2)
I = ∫<0, 2> 2u^2du/(1+u^2)
= 2∫<0, 2> [1 - 1/(1+u^2)] du
= 2[u - arctanu]<0, 2> = 2(2-arctan2)
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