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How do you integrate (ln(x))^2? dx

by maxfails
Tags: integrate, lnx2
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maxfails
#1
Jan29-09, 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.
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tiny-tim
#2
Jan29-09, 10:33 AM
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Quote Quote by maxfails View Post
I'm thinking there's integration by parts involved but not sure.
Hi maxfails!

Yes, use integration by parts with 1 as the function.
fundoo
#3
Jan29-09, 10:34 AM
P: 7
ln xn = t
x = et
dx = et dt

so initial eqn becomes

[tex]\int t^n e^t dt[/tex]

and now integrate by parts

optics.tech
#4
Jan29-09, 01:17 PM
P: 80
How do you integrate (ln(x))^2? dx

ln xn = t
x = et
dx = et dt

so initial eqn becomes

[tex]\int t^n e^t dt[/tex]
[tex]ln \ x^n = t[/tex]

[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...

jostpuur
#5
Jan29-09, 02:32 PM
P: 2,068
Quote Quote by maxfails View Post
it seems you cant use the property ln x^n = n ln x.
The problem is that this formula is [itex]\ln(x^n)=n\ln(x)[/itex], but you are now interested in [itex](\ln(x))^n[/itex] which is different. You probably knew this, but didn't sound very sure about it.

I'm thinking there's integration by parts involved but not sure.
Well tiny-tim of course answered quite sufficiently already, but I thought I would like to say that personally I like writing recursive formulas such as this:

[tex]
(\ln(x))^n = D_x \big(x(\ln(x))^n\big) - n(\ln(x))^{n-1}
[/tex]

fundoo, optics.tech, those were quite confusing comments
NoMoreExams
#6
Jan29-09, 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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