(I)∵f′(x)= x+a x2 ,①a≥0时,f′(x)>0,f(x)在(0,+∞)递增,②a<0时,令f′(x)>0,解得:x>-a,∴f(x)在(-a,+∞)递增,(Ⅱ)a=- 2 时,f′(x)= x? 2 x2 ,f(x)在(1, 2 )递减,在( 2 ,e)递增,∴f(x)min=f( 2 )=ln( 2 )+1= 1 2 ln2+1.