设a∈R,则a=1是f(x)=ln[a+ 2/(x-1)]为奇函数的（充要条件）

当a=1时,有f(x)=ln[1+ 2/(x-1)]=ln[（x+1）/(x-1)]
f(-x)=ln[1+ 2/(-x-1)]=ln[(x-1)/(x+1)]
f(x)+f(-x)=ln[（x+1）/(x-1)]+ln[(x-1)/（x+1）]=ln[（x+1）/(x-1)*(x-1)/（x+1）]
=ln1=0…………充分条件
当f(x)=ln[a+ 2/(x-1)]为奇函数时；有f(x)+f(-x)=0
ln[a+ 2/(x-1)]+ln[a+ 2/(-x-1)]=0
[a+ 2/(x-1)]*[a- 2/(x+1)]=1
a^2-2a/(x+1)+2a/(x-1)-4/(x^2-1)=1
a^2+[2a*(x+1)-2a*(x-1)-4]/(x^2-1)=1
a^2+(4a-4）/(x^2-1)=1
(a^2-1)+4(a-1）/(x^2-1)=0
（a-1）(a+1)+4(a-1）/(x^2-1)=0
(a-1)[(a+1)+4(x^2-1)]=0
a-1=0或 a+4x^2-3=0 恒成立
a=1 或 a=3-4x^2