求函数f(x)=bx/(x^2)-1单调区间（-1<x<1,b不等于0)

f(x)=bx/(x^2)-1
=bx/[(x+1)(x-1)]
=(b/2)·[1/(x+1)+1/(x-1)]
=(b/2)/(x+1)+(b/2)/(x-1)
则f'(x)=-(b/2)/(x+1)^2-(b/2)/(x-1)^2
=-(b/2)·[1/(x+1)^2+1/(x-1)^2]
∵1/(x+1)^2+1/(x-1)^2恒>0,
∴f'(x)的取值只与b有关.

当b>0时,f'(x)=-(b/2)·[1/(x+1)^2+1/(x-1)^2]<0,则函数f(x)在区间（-1,1)上单调递减;
当b<0时,f'(x)=-(b/2)·[1/(x+1)^2+1/(x-1)^2]>0,则函数f(x)在区间（-1,1)上单调递增.