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

by maxfails
Tags: integrate, lnx2
 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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Thanks
P: 26,148
 Quote by maxfails I'm thinking there's integration by parts involved but not sure.
Hi maxfails!

Yes, use integration by parts with 1 as the function.
 P: 7 ln xn = t x = et dx = et dt so initial eqn becomes $$\int t^n e^t dt$$ and now integrate by parts
P: 80
How do you integrate (ln(x))^2? dx

 ln xn = t x = et dx = et 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...

P: 2,068
 Quote by maxfails it seems you cant 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
 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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