f(x)=ln(x+1)-kx/(x+1)(k为常数).

题目

f(x)=ln(x+1)-kx/(x+1)(k为常数).
(1)求f(x)的单调区间
(2)证明不等式:x/ln(x+1)-1

答案

变形:x/ln(x+1)-12
令F(x)=ln(x+1)+4/(x+2)
求导 F'(x)=1/(x+1) -4/(x+2)^2
=x^2/[(x+1)·(x+2)^2]>0
(0