f(x)=(x^2-3x+1)/(x+1) = (x^2+x-4x-4+5)/(x+1) = x - 4 + 5/(x+1)
f'(x) = 1 - 5/(x+1)^2 = { (x+1)^2 - 5 ] / (x+1)^2
x＞1
x∈（1,-1+√5）時(shí),f'(x)＜0,f(x)單調(diào)減
x∈（-1+√5,+∞）時(shí),f'(x)＞0,f(x)單調(diào)增
x=-1+√5時(shí),最小值f(x)min= -1+√5 - 4 + 5/(-1+√5+1) = -5+2√5
值域【-5+2√5,+∞）