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令 √(x+1) = u, 则 x = u^2-1, dx = 2udu
I = ∫<0, 2>[u/(1+u)]2udu = 2∫<0, 2>[u^2/(u+1)]du
= 2∫<0, 2>[(u^2+u-u-1+1)/(u+1)]du
= 2∫<0, 2>[u-1+1/(u+1)]du
= [u^2-2u+2ln(u+1)]<0, 2> = 2ln3
I = ∫<0, 2>[u/(1+u)]2udu = 2∫<0, 2>[u^2/(u+1)]du
= 2∫<0, 2>[(u^2+u-u-1+1)/(u+1)]du
= 2∫<0, 2>[u-1+1/(u+1)]du
= [u^2-2u+2ln(u+1)]<0, 2> = 2ln3
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