woyaolaipi
幼苗
共回答了18个问题采纳率:88.9% 举报
令x=√(e^t-1)
则t=ln(x²+1)
dt=2xdx/(x²+1)
且t=a,x=√(e^a-1)
t=2ln2,x=√3
所以∫[√(e^a-1),√3] [2xdx/(x²+1)]/x=π/6
∫[√(e^a-1),√3] [2dx/(x²+1)]=π/6
2arctanx [√(e^a-1),√3]=π/6
2[π/3-arctan√(e^a-1)]=π/6
arctan√(e^a-1)=π/4
所以√(e^a-1)=1
a=ln2
1年前
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