O(n^2) vs O (n(logn)^2)

Question

Is time complexity O(n^2) or O (n(logn)^2) better?

I know that when we simplify it, it becomes

O(n) vs O((logn)^2)

Answer 1

(logn)^2 is also < n.

Take an example:

 n = 5
 log n = 0.6989....
 (log n)^ 2 = 0.4885..

You can see, (long n)^2 is further reduced.

Even if you take any bigger value of n e.g. 100,000,000 , then

   log n = 9
   (log n)^ 2 = 81

which is far less than n.