已知 f[(1-x)/(1+x)]=(1-x^2)/(1+x^2)

令(1-x)/(x+1)=t则1-x=tx+t→x=(1-t)/(t+1)∴f(t)=[1-(1-t)^2/(t+1)^2]/[1+(1-t)^2/(t+1)^2]上下都乘以(T+1)^2得[(t+1)^2-(t-1)^2]/[(t+1)^2+(t-1)^2]=4t/(2t^2+2)=2t/(t^2+1)→f(x)=2x/(x^2+1)