为自己加油

计算∫₀¹[1/(1+e^x)]dx．

1-ln(1+e)+ln2 分析 ∫₀¹[1/(1+e^x)]dxfoI=∫₀¹[(1+e^x-e^x)/(1+e^x)]dx=∫₀¹[1-e^x/(1+e^x)]dx =∫₀¹dx-∫₀¹[1/(1+e^x)]d(1+e^x) =(-ln(1+e))|₀¹=1-ln(1+e)+ln2．