令t=x+1
f(t)=(lnt)^2-(t-1)^2/t=(lnt)^2-(t-2+1/t)
则f'(t)= 2(lnt)/t-1+1/t^2=1/t*[ 2lnt-t+1/t]
g(t)=2lnt-t+1/t
g'(t)=2/t-1-1/t^2=-(1/t-1)^2