Question

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

Answer 1

`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`