If we apply Master’s Theorem:
$=> n^2logn = n^{log_24}$
$=> n^2logn = n^2$ $… (1)$
In essence, $n^2logn$ is not polynomially greater than $n^2$
Hence, we apply a more generalized Master’s Theorm which says:
If $f(n) = \theta(n^{log_ba}log^kn)$ where $k>0:$
$ => T(n) = \theta(n^{log_ba} \cdot log ^{k+1}n)$
Let me make it little simpler here, it means where you see the log term, multiply again with the log term.
In our case it was:
$=> \mathbf{n^2logn} = n^2$ … From $(1)$
Hence we will multiply $logn$ again to LHS term
Our answer becomes:
$ => \mathbf{n^2logn} \cdot logn$
$ => \mathbf{n^2log^2n}$
