定义域为R,
令t=x^2>=0
则f=(t+1)/(t^2+1)=t/(t^2+1)+1/(t^2+1)
t=0时,f=1
t>0时,f=1/(t+1/t)+1/(t^2+1)
因为t+1/t>=2,故0∣f(x)∣=∣(x^2+1)/(x^4+1)∣≤(x^2+1)^2/(x^4+1）=(x^4+1+2x^2)/(x^4+1)=1+2x^2/(x^4+1)≤1+1=2故f(x)在R内有界书上是这样写的，1+2x^2/(x^4+1)≤1+1 这一步没看懂这是用公式：2ab<=a^2+b^2a=x^2, b=1因此有：2x^2<=x^4+12x^2/(x^4+1)≤11+2x^2/(x^4+1)≤1+1