f'(x)=x+1/x+(a-4)
f(x)在(1,+∞)上是增函数
所以x＞1时,f'(x)=x+1/x+(a-4)≥0
即x^2+(a-4)x+1≥0
即y=x^2+(a-4)x+1与x轴最多有1个交点,
△=(a-4)^2-4=a^2-8a+12≤0
2≤a≤6