We have
#I = int (lnx)^2/x^2dx#
If #z = lnx# then #dz = dx/x# and
#z = ln x# so #e^z = x#
#I = int z^2e^(-z)dz#
Integrating by parts
#u = z^2# so #du = 2z dz#
#dv = e^(-z) dz# so #v = -e^(-z)#
#I = u*v - int v du#
#I = -z^2e^(-z) + 2intze^(-z)dz#
Integrating by parts once again
#u = z# so #du = dz#
#dv = e^(-z)dz# so #v = -e^(-z)#
#I = -z^2e^(-z) + 2(-ze^(-z) + inte^(-z)dz)#
#I = -z^2e^(-z) - 2ze^(-z) - 2e^(-z) + c#
#I = -e^(-z)(z^2 + 2z + 2) + c#
But we want to have an answer in terms of #x# so if we remember that #z = ln(x)# we'll have
#I = -e^(-ln(x))((lnx)^2 + 2lnx + 2) +c#
#I = c-((lnx)^2 + 2lnx + 2)/x#