h'(x)=(2x+b)e^x+(x^2+bx+1)e^x=[x^2+(b+2)x+b+1]e^x
令h'(x)=0,得x=-1或x=-(b+1)
当-1>-(b+1),即b>0时,h(x)的递减区间是[-(b+1),-1]
当-1=-(b+1),即b=0时,h'(x)>=0恒成立,故此时h(x)在R上递增
当-1