f'(x)=e^x- 1+x

Obviously, when x>0 has f' (x) > 0; 0, when x

Therefore, the monotonically increasing interval of f(x) is (0,+infinity), and the monotonically decreasing interval is (-∞, 0).

f(x)> 1/2x^2+ax

It is equivalent to the constant establishment of e x-(a+ 1) x > 0.

Let g (x) = e x-(a+1) x.

g'(x)=e^x-a- 1

If a+ 1

If a+ 1=0, it obviously meets the problem.

If a+1>; 0, there is a minimum value when e x = a+ 1, so let e t = a+ 1.

G (t) = (a+1) (1-t) > 0, so t

So a+ 1 = e t

a & lte- 1

finally

The range of a is [- 1, e- 1].