f(x1)-f(x2)=x1+√(x1^2+1)-x2-√(x2^2+1)
=(x1-x2)+√(x1^2+1)-√(x2^2+1)
=(x1-x2)+[√(x1^2+1)-√(x2^2+1)][√(x1^2+1)+√(x2^2+1)]/[√(x1^2+1)+√(x2^2+1)]
=(x1-x2)+(x1^2+1-x2^2-1)/[√(x1^2+1)+√(x2^2+1)]
=(x1-x2)+(x1^2-x2^2)/[√(x1^2+1)+√(x2^2+1)]
=(x1-x2)+(x1-x2)(x1+x2)/[√(x1^2+1)+√(x2^2+1)]
=(x1-x2){1+(x1+x2)/[√(x1^2+1)+√(x2^2+1)]}
设任意的x1>x2>0,则(x1-x2)>0,{1+(x1+x2)/[√(x1^2+1)+√(x2^2+1)]}>0
则f(x1)-f(x2)>0,即单调递增.
设任意的0≥x1>x2,
[√(x1^2+1)+√(x2^2+1)]>-（x1+x2）
则(x1+x2)/[√(x1^2+1)+√(x2^2+1)]>-1
则{1+(x1+x2)/[√(x1^2+1)+√(x2^2+1)]}>0
则f(x1)-f(x2)>0,即单调递增.
综上所述：在（ -∞ +∞）上单调递增.