Is the derivative of (ln x)^2 a u-sub?

[amb1989]

Is this a u-sub? I went through and got 1/(2x)^2 but I am not sure if that is correct.

 

[danago]

You can use a substitution, but the answer is not quite 1/(2x)^2.

How did you arrive at that answer?

 

[amb1989]

You can use a substitution, but the answer is not quite 1/(2x)^2.

How did you arrive at that answer?

Actually I asked the wrong question.

I'm working on an integration by parts problem it asks me to integrate what is in the topic title.
I went about it by saying that U = ln(x)^2 and that dv = 1.

When I went through and plugged everything into the integration by parts formula I arrived at

xln(x)-(integral sign)(1)(1/(2x)^2)

The thing that I am getting hooked up is trying to take that integral of 1/(2x)^2. I think I messed up somewhere but I am not sure where to look. Any suggestions?

 

[danago]

Hmm I am not really sure how you ended up xln(x)-(integral sign)(1)(1/(2x)^2).

I would choose the same parts as you did, i.e. u = (ln x)^2 and dv = 1, but I am not sure you applied the formula correctly.

\int u dv =uv - \int v du

Is that what you are using?

 

[amb1989]

Hmm I am not really sure how you ended up xln(x)-(integral sign)(1)(1/(2x)^2).

I would choose the same parts as you did, i.e. u = (ln x)^2 and dv = 1, but I am not sure you applied the formula correctly.

\int u dv =uv - \int v du

Is that what you are using?

Yeah that's what I am using. du = 1/(2x^2) v= x

(lnx)^2)(x)-(integral)(1/(2x)^2)(1)

Did I take the derivative of (ln(x))^2 wrong?

 

[danago]

Yea maybe have another look at the derivative of (ln x)²

You could use the substitution z = ln(x), or treat it as (ln x)² = (ln x)(ln x) and apply the product rule.