T(n)=2T(n/2)+n/log n comparing with T(n)=aT(n/b)+f(n) where b>1 f(n) is positive
applying master mathed-
nlog22 =n > f(n) so probably case1 applicable ....hence the solution is ⊖(n)
but some times the gap between f(n) and nlogb a is small and it is log n in that that case the solution shifted with one log value. but here log ≍ √n hence f(n) is approximate n/√n=√n hence final solution is ⊖(n)