y=f(x)=(2^x+1)/(2^x-1)
=> 2^x+1=y*2^x-y
=> (y-1)*2^x=y+1
=> 2^x=(y+1)/(y-1)
=> x=log(2) [(y+1)/(y-1)]
故原函数的反函数为
y=log(2)[(x+1)/(x-1)]
即以2为底数,[(x+1)/(x-1)]的对数