As the function is continuous in the closed intervals #x in[1/2,1)# and #x in (1,2]# evaluate:
#int_(1/2)^2 dx/(x(lnx)^4) = int_(1/2)^1 dx/(x(lnx)^4) + int_1^2 dx/(x(lnx)^4) #
Change #t= 1/x# in the second integral:
# int_1^2 dx/(x(lnx)^4) = int_1^(1/2) -t/t^2 dt/(ln(1/t))^4#
exchange the limits of integration:
# int_1^2 dx/(x(lnx)^4) = int_(1/2)^1 dt/(t(-ln(t))^4)#
# int_1^2 dx/(x(lnx)^4) = int_(1/2)^1 dt/(t(ln(t))^4)#
Then:
#int_(1/2)^2 dx/(x(lnx)^4) = 2int_(1/2)^1 dx/(x(lnx)^4)#
#int_(1/2)^2 dx/(x(lnx)^4) = 2int_(1/2)^1 (d(lnx))/(lnx)^4#
#int_(1/2)^2 dx/(x(lnx)^4) = [-2/(3(lnx)^3)]_(1/2)^1 #
#int_(1/2)^2 dx/(x(lnx)^4) = 2/(3(ln(1/2))^3)-2/3 lim_(x->1^-) 1/(lnx)^3 = +oo#
