设u=
x+1
x-1,任取x2>x1>1,则
u2-u1=
x2+1
x2-1-
x1+1
x1-1
=
(x2+1)(x1-1)-(x1+1)(x2-1)
(x2-1)(x1-1)
=
2(x1-x2)
(x2-1)(x1-1).
∵x1>1,x2>1,∴x1-1>0,x2-1>0.
又∵x1<x2,∴x1-x2<0.
∴
2(x1-x2)
(x2-1)(x1-1)<0,即u2<u1.
当a>1时,y=logax是增函数,∴logau2<logau1,
即f(x2)<f(x1);
当0<a<1时,y=logax是减函数,∴logau2>logau1,
即f(x2)>f(x1).
综上可知,当a>1时,f(x)=loga
x+1
x-1在(1,+∞)上为减函数;
当0<a<1时,f(x)=loga
x+1
x-1在(1,+∞)上为增函数.