计算∫01[1/(1+ex)]dx.【正确答案】:1-ln(1+e)+ln2 分析 ∫01[1/(1+ex)]dxfoI=∫01[(1+ex-ex)/(1+ex)]dx=∫01[1-ex/(1+ex)]dx =∫01dx-∫01[1/(1+ex)]d(1+ex) =(-ln(1+e))|01=1-ln(1+e)+ln2.
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