f(x)=ax/(x^2-1)
f(-x)=-xa/(x^2-1)
f(x)=-f(x)
故f(x)在x∈(-1,1)内为奇函数
f(1/2)=-2a/3
f(0)=0
当a>0时,f(1/2)0,故f(x)在x∈(-1,1)内单调减
当af(0),1/2>0,故f(x)在x∈(-1,1)内单调增