∫[0→2] f(x-1)dx
令x-1=u,则dx=du,u：-1→1
=∫[-1→1] f(u)du
=∫[-1→0] f(u)du+∫[0→1] f(u)du
=∫[-1→0] 1/(1+e^(u+1))du+∫[0→1] 1/(1+u) du
=∫[-1→0] e^(u+1)/[e^(u+1)(1+e^(u+1))]du+ln(u+1) |[0→1]
=∫[-1→0] 1/[e^(u+1)(1+e^(u+1))]d(e^(u+1))+ln2
=∫[-1→0] 1/e^(u+1)d(e^(u+1))-∫[-1→0] 1/(1+e^(u+1))d(e^(u+1))+ln2
=ln[e^(u+1)]-ln[e^(u+1)+1]+ln2 |[-1→0]
=1-ln(e+1)+ln2+ln2
=1+2ln2-ln(e+1)