求∫（上限+∝下限0）1/(1+e^x)dxln2

Question

求∫（上限+∝ 下限0 ）1/(1+e^x) dx
ln2

Answer 1

∫ 1/(1 + e^x) dx
= ∫ (1 + e^x - e^x)/(1 + e^x) dx
= ∫ dx - ∫ e^x/(1 + e^x)
= x - ∫ d(e^x + 1)/(1 + e^x)
= ln(e^x) - ln(1 + e^x) + C
= ln[e^x/(1 + e^x)] + C
= ln[1/(1 + e^-x)] + C
= - ln(1 + e^-x) + C
∫(0~∞) 1/(1 + e^x) dx
= - ln(1 + e^-x) |(0~∞)
= - [ln(1 + 0) - ln(1 + 1)]
= - [0 - ln2]
= ln2