用换元法思路很直接
令t=(1-x)/(1+x)
(1+x)t=1-x
tx+x=1-t
x=(1-t)/(1+t)
f(t)
=(1-x^2)/(1+x^2)
=[1-(1-t)^2/(1+t)^2]/[1+(1-t)^2/(1+t)^2]
=[(1+t)^2-(1-t)^2]/[(1+t)^2+(1-t)^2]
=[(1+2t+t^2)-(1-2t+t^2)]/[(1+2t+t^2)+(1-2t+t^2)]
=(4t)/(2+2t^2)
=2t/(1+t^2)
将t换回x,即得
f(x)=2x/(1+x^2)