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:

$$(\ln(x))^n = D_x \big(x(\ln(x))^n\big) - n(\ln(x))^{n-1}$$

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).