The recursion tree has log n levels, each contributing n/log(n/2^i). Level 0: n/log n, Level 1: n/log(n/2), ..., Level log n: O(log n). Using the integral approximation, the sum of n/log(n/2^i) for i=0 to log n is O(n log log n). This is because we're essentially summing 1/log(x) which gives the log log factor.