First, we will use substitution
Let t = ln(x) => dt = 1/xdxt=ln(x)⇒dt=1xdx and x = e^tx=et
Then
intln^2(x)/x^2dx = intln^2(x)/x*1/xdx = intt^2e^-tdt∫ln2(x)x2dx=∫ln2(x)x⋅1xdx=∫t2e−tdt
Next, we will use the integration by parts forumla
intudv = uv-intvdu∫udv=uv−∫vdu
Integration by Parts 1:
Let u = t^2u=t2 and dv = e^-tdtdv=e−tdt
Then du = 2tdu=2t and v = -e^-tv=−e−t
Applying the formula:
intt^2e^-tdt = -t^2e^-t+2intte^-tdt∫t2e−tdt=−t2e−t+2∫te−tdt
Integration by Parts 2:
Focusing on the remaining integral...
Let u = tu=t and dv = e^-tdtdv=e−tdt
Then du = dtdu=dt and v = -e^-tv=−e−t
Applying the formula:
intte^-tdt = -te^-t + inte^-tdt∫te−tdt=−te−t+∫e−tdt
=-te^-t-e^-t+C=−te−t−e−t+C
=-e^-t(t+1)+C=−e−t(t+1)+C
Substituting back, we have
intt^2e^-tdt = -t^2e^-t+2[-e^-t(t+1)]+C∫t2e−tdt=−t2e−t+2[−e−t(t+1)]+C
=-e^-t(t^2+2t+2)+C=−e−t(t2+2t+2)+C
Finally, substituting xx back in gives our final result:
intln^2(x)/x^2dx = -(ln^2(x)+2ln(x)+2)/x+C∫ln2(x)x2dx=−ln2(x)+2ln(x)+2x+C
