let
u=x-1
∫(0->2) f(x-1) dx
=∫(-1->1) f(u) du
=∫(-1->0) f(u) du +∫(0->1) f(u) du
=∫(-1->0) du/(1+e^u) +∫(0->1) du/(1+u)
t = e^u
=∫(e^(-1)->1) dt/[t(1+t)] +[ ln|1+u| ]|(0->1)
=∫(e^(-1)->1) [1/t-1/(1+t)] dt+ ln2
=[ln|t/(1+t)|]|(e^(-1)->1) + ln2
= ln(1/2) - ln[ (1/e)/(1+ 1/e) ] + ln2
=1 - ln|1+1/e|
=ln|1+e|