证:f(x-1)=ln(x-1)-1/(x-1)
f(x-1)-2x+5=ln(x-1)-1/(x-1)-2(x-1)+3=lnt-1/t-2t+3(t=x+1,t≥1)
设g(t)=lnt-1/t-2t+3(t≥1),g'(t)=1/t+1/t^2-2在t≥1时恒小于0,g(t)单调递减
g(t)≤g(t)max=g(1)=0
即f(x-1)≤2x－5