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I am suppose to find $\int(\ln x)^3$ by using the proof (I have to prove this part first) of

$$\int (\ln x)^n dx = x(\ln x)^n - n \int (\ln x)^{n-1} dx$$

I can not prove it and I do not know how to work with n powers like that, to me it doesn't even look right.

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To prove the last formula (do you see how to apply it?), try integration by parts. The $u$ and $dv$ that work are somewhat surprising; for now let me just say that if your $dv$ is hard to integrate, then you're doing too much work. – Dylan Moreland Jun 8 '12 at 19:21
I do not understand how to do it with n as a power though, it doesn't make sense. – user138246 Jun 8 '12 at 19:25
@Jordan: You have seen formulas like $\frac{d}{dx}x^n=nx^{n-1}$? Does that make sense? – Jonas Meyer Jun 8 '12 at 19:26
Would it be easier to think of the case $n=3$, at first? In essence it is no easier than the general case, but the concreteness might help. – Dylan Moreland Jun 8 '12 at 19:30
(Joke, sort of): If one is asked to prove an explicit integration formula, one way is to differentiate the "answer." – André Nicolas Jun 8 '12 at 20:27

We have

$$\mathrm I(n)=\int (\log x)^ndx$$

integrate by parts with

$$dx=du$$ $$\log^n x = v$$

$$x=u$$ $$n (\log x)^{n-1} \frac{1}{x}dx=dv$$

Spoiler ahead

$$\mathrm I(n)=x\log^n x-\int n (\log x)^{n-1}\dfrac{1}{x}x dx$$

$$\mathrm I(n)=x\log^n x-n\int (\log x)^{n-1} dx$$

$$\mathrm I(n)=x\log^n x-n \mathrm I (n-1)$$

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You are using ln and log as the same thing but they are not the same, does this not change the answer? – user138246 Jun 8 '12 at 19:34
@Jordan Many mathematicians use $\log$ instead of $\ln$ for the base $e$ logarithm. This is probably because it can be considered "the" logarithm. – Argon Jun 8 '12 at 19:35
I is the integral? I do not really follow what happened in the last step. – user138246 Jun 8 '12 at 19:42
"$I(n)$" means "that integral, with an n in the exponent". So "$I(n - 1)$" means "that integral, with an $n - 1$ in the exponent"... – The Chaz 2.0 Jun 8 '12 at 19:52
Oh so it means the integrand to the n power? So with this formula does that mean that the power just reduces by one each time I use the formula? – user138246 Jun 8 '12 at 19:59

Integration by parts is a useful technique. You can use $u=(\ln x)^n$ and $v'=1$ in the formula $\int uv' dx = uv-\int vu' dx$.

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