∫1/[e^x (1+e^x)]dx ＝∫[1/e^x－1/(1＋e^x)]dx ＝∫[e^(-x)－e^(-x)/(1＋e^(-x))]dx ＝－∫e^(-x) d(-x) ＋∫1/(1＋e^(-x))] d(1+e^(-x)) ＝－e^(-x)＋ln[(1＋e^(-x))]＋C