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I know that it is an integration by parts, but I don't know what to set as my u and dv.

ILATE

L = logarithmic
I = inverse trigonometric
A = algebraic
T = trigonometric
E = exponential

I>L>A>T>E
That is the order of priority.

would I set (lnx)^2 as my U. does it become part of the log family or exponential ?

give it a try, you would never know if it works or not without trying

You have to consider the inside function. f(x)^2
f(x) belongs to?

yes that is why I put lnx as my u

lmltfy!
\[\int 3x(ln(x))^2 \ dx\]

yes that is why I put lnx as my u|dw:1440859979280:dw|

This is what I did. Does it look correct?

It is too blur to see but I think for uv, you are ok, for \(\int vdu\) it is not ok to me.

You let u = (ln(x))^2 --> \(du = \dfrac{2lnx}{x}dx\)
dv = 3xdx --> \(v = \dfrac{3x^2}{2}\) , right?

apply integral by part again for the second term, you can get the answer.

I backtracked and fixed it.

"When you can take the pebble from my hand, it will be time for you to leave."