如题所述
原式=(1/2)∫[(x-1)/(x 1)^2 1/(x-1)]dx
=(1/2)∫[(x 1-2)/(x 1)^2 1/(x-1)]dx
=(1/2)∫[1/(x 1)-2/(x 1)^2 1/(x-1)]dx
=(1/2)[ln(x 1) 2/(x 1) ln(x-1)] C
=(1/2)[ln(x 1) ln(x-1)] 1/(x 1) C