I'm trying to figure out how you would apply the Master Theorem to these two cases: $$ \begin{split} T(n) &= 2 T(n/2) + n \log(n) + 3n\\ T(n) &= 2 T(n/2) + n (\log(n))^2 \end{split} $$
I have some sort of idea for the first one, since you could say that $f(n) = n \log(n) + 3n$ and then since $n\log(n)$ grows fastest, the rest of it can be dropped which gives $T(n) = 2 T(n/2) + n\log(n)$ but I've no idea on how to approach the second one at all. I know how to apply the Master Theorem, I'm just not sure how to apply it to these cases since they're not in the usual form.
HINT
Putting this into the form $T(n) = aT(n/b) + f(n)$, we have $a=b=2$ and $f(n) = n \ln^2(n)$, so $c = \log_b a = \log_2 2 = 1$, and $$ f(n) = n \ln^2(n) = \Theta\left(n^c \ln^{1+1}n\right), $$ This means you are in the second case of the Master Theorem in Wikipedia formulation, for $k=1$. Can you now complete the problem?
UPDATE
The conclusion of the master theorem is that $$ T(n) = \Theta \left(n^c \log^{k+1} n\right), $$ but we already saw that $c=1$ and $k=1$, from the formulation above. Can you plug in?