若lnx/(x+1) +1/x > lnx/(x-1)+k/x 对于x>o且不等于1 恒成立,求k取值范围

lnx/(x+1) +1/x > lnx/(x-1)+k/x lnx-ln(x+1)-lnx+ln(x-1)>(k-1)/xln[(x-1)/(x+1)]>(k-1)/x ,因为x>o,所以,x ln[(x-1)/(x+1)]>k-1对于任意 x>o,(x-1)/(x+1)﹤1则,ln[(x-1)/(x+1)]﹤0x ln[(x-1)/(x+1)]﹤0故k-1﹤0k﹤...

上面这个人乱做的.