Is this a u-sub? I went through and got 1/(2x)^2 but I am not sure if that is correct.
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
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?
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.
[tex]\int u dv =uv - \int v du[/tex]
Is that what you are using?
Yeah that's what I am using. du = 1/(2x^2) v= x
Did I take the derivative of (ln(x))^2 wrong?
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.
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