First, we will use substitution
Let #t = ln(x) => dt = 1/xdx# and #x = e^t#
Then
#intln^2(x)/x^2dx = intln^2(x)/x*1/xdx = intt^2e^-tdt#
Next, we will use the integration by parts forumla
#intudv = uv-intvdu#
Integration by Parts 1:
Let #u = t^2# and #dv = e^-tdt#
Then #du = 2t# and #v = -e^-t#
Applying the formula:
#intt^2e^-tdt = -t^2e^-t+2intte^-tdt#
Integration by Parts 2:
Focusing on the remaining integral...
Let #u = t# and #dv = e^-tdt#
Then #du = dt# and #v = -e^-t#
Applying the formula:
#intte^-tdt = -te^-t + inte^-tdt#
#=-te^-t-e^-t+C#
#=-e^-t(t+1)+C#
Substituting back, we have
#intt^2e^-tdt = -t^2e^-t+2[-e^-t(t+1)]+C#
#=-e^-t(t^2+2t+2)+C#
Finally, substituting #x# back in gives our final result:
#intln^2(x)/x^2dx = -(ln^2(x)+2ln(x)+2)/x+C#
