Making #y = log_e x# and
#dy = dx/x#
substituting in #int (log_e x)^n dx# we have
#int (log_e x)^n dx equiv int y^n e^y dy#
Now
#d/(dy)(y^n e^y) = ny^{n-1}e^y+y^n e^y#
so
#int y^n e^y dy = y^n e^y-n int y^{n-1} e^y dy + c_1#
or calling #I_n = int y^n e^y dy# we have
#I_n = y^n e^y -n I_{n-1} + c_1# or supposing #c_1=0#
#I_n+nI_{n-1} = y^n e^y# Now making #J_n=e^yI_n#
#J_n+nJ_{n-1} = y^n#
This recurrence equation has the solution
https://en.wikipedia.org/wiki/Recurrence_relation
#J_n = (-1)^n n!(c_0+sum_{k=0}^{n-1}((-1)^k)/((k+1)!)y^{k+1})#
then
#I_n = (c_0+sum_{k=0}^{n-1}((-1)^{n+k} n!)/((k+1)!)(log_e x)^{k+1})x#
so
#int (log_e x)^n dx = (c_0+sum_{k=0}^{n-1}((-1)^{n+k} n!)/((k+1)!)(log_e x)^{k+1})x#
