$T(n)=2T(n/2) + n\log^2(n)$.If I try to substitute $m = \log(n)$ I end up with $T(2^m)=2 T(2^{m-1}) + 2^m\log^{2}(2^m)$.Which isn't helpful to me. Any clues?PS. hope this isn't too localized. I specified that the problem was a squared logarithm which should make it possible to find for others wondering about the same thing.
Solving recurrence with logarithm squared $T(n)=2T(n/2) + n \log^2n$
asymptotics;proof techniques;recurrence relation
This is indeed the second case in the Master Theorem. For the standard recursion form $$T(n)=a\;T(n/b)+f(n),$$you get $a=b=2$, and therefore  $f(n)=\Theta(n^{\log_b a} \log^2 n)=\Theta(n  \log^2 n)$.Applying the Master theorem yields $T(n)=\Theta(n\log^3 n)$.