设x1 > x2,f(x1) - f(x2) = x1^3 + 1 - (x2^3 + 1) = x1^3 - x2^3 = (x1 - x2)(x1^2 + x1x2 + x2^2) = (x1 - x2)((x1 + 1/2x2)^2 + 3/4x2^2) > 0,即f(x1) > f(x2),所以函数f(x)在r上单调递增