Assume that I have a data sets witch consists of $n$ elements. Then, I apply two algorithms to this set, one after another, and each algorithm has time complexity $O(n^2\log{n})$. What is the oeverall complexity?
My approach is the following. From the one hand side, normally, from what I saw, if the complexity is additive, the we choose the worst one, i.e. $$ O(\alpha_{n}) + O(\beta_{n}) = O(\alpha_{n}), $$ if $\alpha_{n}/\beta_{n} \to\infty$, as $n\to\infty$.
At the same time, in our case we have that $$ n^2\log{n} + n^2\log{n} = 2n^2\log{n} = n^2\log{n^{2}} $$ Obviously, $\log{n}$ and $\log{n^{2}}$ are different.
What is the solution to my confusion?