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How do you integrate (ln(x))^2? dx 
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#1
Jan2909, 10:19 AM

P: 11

it seems you cant use the property ln x^n = n ln x.
I'm thinking there's integration by parts involved but not sure. 


#2
Jan2909, 10:33 AM

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Thanks
P: 26,148

Yes, use integration by parts with 1 as the function. 


#3
Jan2909, 10:34 AM

P: 7

ln x^{n} = t
x = e^{t} dx = e^{t} dt so initial eqn becomes [tex]\int t^n e^t dt[/tex] and now integrate by parts 


#4
Jan2909, 01:17 PM

P: 80

How do you integrate (ln(x))^2? dx
[tex]e^{ln \ x^n} = e^t[/tex] [tex]x^n=e^t[/tex] [tex]\frac{d}{dt} \ (x^n)=\frac{d}{dt} \ (e^t)[/tex] [tex]0=e^t[/tex] remember that: [tex]x=exp \ y \Leftrightarrow y=ln \ x[/tex] So [tex]0=e^t \Leftrightarrow t = ln \ 0[/tex] Since ln 0 is undefined, so t is undefined too... 


#5
Jan2909, 02:32 PM

P: 2,068

[tex] (\ln(x))^n = D_x \big(x(\ln(x))^n\big)  n(\ln(x))^{n1} [/tex] fundoo, optics.tech, those were quite confusing comments 


#6
Jan2909, 02:32 PM

P: 626

Umm there's a difference between (ln(x))^2 and ln(x^2). The first is what you seem to have, the latter is 2*ln(x).



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