令u = 1 + √(4 - x)，4 - (u - 1)² = x，- 2(u - 1) du = dx
当x = 0，u = 3
当x = 3，u = 2
∫(0→3) 1/[1 + √(4 - x)] dx
= ∫(3→2) 1/u · - 2(u - 1) du
= 2∫(2→3) (u - 1)/u du
= 2∫(2→3) (1 - 1/u) du
= [2u - 2ln(u)]：(2→3)
= [2(3) - 2ln(3)] - [2(2) - 2ln(2)]
= 6 - 2ln(3) - 4 + 2ln(2)
= 2 + 2ln(2/3)
令√(4-x)=u，则x=4-u²，dx=-2udu，u：2→1
∫[0→3] 1/[1+√(4-x)]dx
=∫[2→1] -2u/(1+u) du
=2∫[1→2] (u+1-1)/(1+u) du
=2∫[1→2]1du-2∫[1→2] 1/(1+u) du
=2u-2ln(u+1) |[1→2]
=4-2ln3-2+2ln2
=2+2ln2-2ln3
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