已知函数f(x)=alnx+(a+1)/2x^2+1讨论函数f(x)的单调性

f(x)=alnx+(a+1)/2·x²+1,x＞0
f'(x)=a/x+(a+1)x=[(a+1)x²+a]/x
(1) a=-1,f'(x)=-1/x＜0,f(x)在(0,+∞)单调减；
(2) a＜-1,(a+1)x²+a＜0恒成立,f'(x)＜0,f(x)在(0,+∞)单调减；
(3) -1＜a＜0,令f'(x)=0,即 考虑到x＞0,解得x=√[-a/(a+1)],
在(0,√[-a/(a+1)]),f'(x)＜0,f(x)单调减;
在(√[-a/(a+1)],+∞),f'(x)＞0,f(x)单调增;
(4)a≧0,(a+1)x²+a≧0恒成立,f'(x)≧0,f(x)在(0,+∞)单调增.
综上,.