答：
∫1/[1+e^(x-1)] dx
=∫[1+e^(x-1)-e^(x-1)]/[1+e^(x-1)] dx
=∫1-e^(x-1)/[1+e^(x-1)] dx
=x-∫1/[1+e^(x-1)] d[1+e^(x-1)]
=x-ln[1+e^(x-1)] + C
所以定積分∫(0到1)1/[1+e^(x-1)] dx
=x-ln[1+e^(x-1)] |(0到1)
=1-ln(1+e^0)-0+ln(1+e^(-1))
=1-ln(2+2/e)