# Homework Help: The derivative of (ln(x))^2

1. Oct 8, 2009

### amb1989

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

2. Oct 8, 2009

### danago

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

How did you arrive at that answer?

3. Oct 8, 2009

### amb1989

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?

4. Oct 8, 2009

### danago

Hmm im 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 im not sure you applied the formula correctly.

$$\int u dv =uv - \int v du$$

Is that what you are using?

5. Oct 8, 2009

### amb1989

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?

6. Oct 8, 2009

### 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.