What is the antiderivative of # (ln x)^2/x^2#?

1 Answer
Mar 17, 2016

Answer:

#intln^2(x)/x^2dx = -(ln^2(x)+2ln(x)+2)/x+C#

Explanation:

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#