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

[maxfails]

it seems you can't use the property ln x^n = n ln x.

I'm thinking there's integration by parts involved but not sure.

 

[tiny-tim]

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]

ln xⁿ = t
x = e^t
dx = e^t dt

so initial eqn becomes

\int t^n e^t dt

and now integrate by parts

 

[optics.tech]

ln xⁿ = t
x = e^t
dx = e^t dt

so initial eqn becomes

\int t^n e^t dt

ln \ x^n = t

e^{ln \ x^n} = e^t

x^n=e^t

\frac{d}{dt} \ (x^n)=\frac{d}{dt} \ (e^t)

0=e^t

remember that:

x=exp \ y \Leftrightarrow y=ln \ x

So

0=e^t \Leftrightarrow t = ln \ 0

Since ln 0 is undefined, so t is undefined too...

 

[jostpuur]

it seems you can't use the property ln x^n = n ln x.

The problem is that this formula is \ln(x^n)=n\ln(x), but you are now interested in (\ln(x))^n 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:

<br /> (\ln(x))^n = D_x \big(x(\ln(x))^n\big) - n(\ln(x))^{n-1}<br />

fundoo, optics.tech, those were quite confusing comments

 

[NoMoreExams]

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).