Question

求1/[1+e^(x-1)]在(0,1)上的定积分,

Answer 1

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

所以定积分∫(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)