f'(x)=(1/x)+(a/x²)=(ax+1)/(x²)
ax+1=0的根是x=-1/a
(1)若-1/a≤1,即:a≥0或a≤-1时,此时f'(x)≥0,即此时函数在这个区间内递增;
(2)若-1/a≥e,即:-1/e≤a