函数y=[(2^x-1)/(2^x+1)]+ln(x-1)/(x+1)是偶函数还是奇函数?

f(x)=y=[(2^x-1)/(2^x+1)]+ln(x-1)/(x+1)
f(-x)=[2^(-x)-1]/[2^(-x)+1]+ln(-x-1)/(-x+1)
[2^(-x)-1]/[2^(-x)+1]
上下同乘2^x
=(1-2^x)/(1+2^x)
=-(2^x-1)/(2^x+1)
ln(-x-1)/(-x+1)=ln[-(x+1)/-(x-1)]=ln(x+1)/(x-1)=ln{1/[(x-1)/(x+1)]}
=ln[(x-1)/(x+1)]^-1
=-ln(x-1)/(x+1)
所以f(-x)=-(2^x-1)/(2^x+1)-ln(x-1)/(x+1)
=-[(2^x-1)/(2^x+1)+ln(x-1)/(x+1)]
=-f(x)
又定义域
(x-1)/(x+1)>0
所以(x-1)(x+1)>0
x>1,x