f（跟号x+1）=x+2倍根号x,求f(x+1),f(x的平方）注：根号x+1,根号不包括1

换元法
令t=√x+1
√x=t-1,t>=1
x=(t-1)^2
f(t)=(t-1)^2+2(t-1)
=t^2-1
f(x)=x^2-1,x>=1
f(x+1)=(x+1)^2-1
=x^2+2x,x>=0,
f(x^2)=((x^2))^2-1=x^4-1,|x|>=1