计算定积分∫[4,1]dx/x+√x

令t = √x,t² = x,2t dt = dx
x = 1,t = 1；x = 4,t = 2
∫(1→4) dx/(x + √x)
= ∫(1→2) 2t/(t² + t) · dt
= ∫(1→2) 2/(1 + t) dt
= 2[ln(1 + t)]：(1→2)
= 2ln(1 + 2) - 2ln(1 + 1)
= 2ln(3/2)
= ln(9/4)
再问： 2ln(1 + 2) - 2ln(1 + 1) 这一步是怎么变成 = 2ln(3/2) ？
再答： 对数公式lnA - lnB = ln(A/B)