1. Reduction Forumula

I'm trying to prove:

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

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

It's not obvious what substition I should use here. If I let $\displaystyle u=(ln(x))^{n-1}$ and $\displaystyle dv=ln(x)dx$ then I have $\displaystyle du=(n-1)(ln(x))^{n-2}\frac{1}{x}dx$ and $\displaystyle v=xln(x)-x$.

This turn into a really complicated thing, and I can't see were I will be able to find $\displaystyle I$ in this mess. I don't really expect anyone to prove this for me, since it's too much work. However, I just need some guidance. Is there some trick I'm not noticing?

I'm trying to prove:

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

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

It's not obvious what substition I should use here. If I let $\displaystyle u=(ln(x))^{n-1}$ and $\displaystyle dv=ln(x)dx$ then I have $\displaystyle du=(n-1)(ln(x))^{n-2}\frac{1}{x}dx$ and $\displaystyle v=xln(x)-x$.

This turn into a really complicated thing, and I can't see were I will be able to find $\displaystyle I$ in this mess. I don't really expect anyone to prove this for me, since it's too much work. However, I just need some guidance. Is there some trick I'm not noticing?
use integration by parts with

$\displaystyle u=[\ln(x)]^n \implies du =\frac{n[\ln(x)]^{n-1}}{x}dx$
$\displaystyle dv=dx \implies v=x$

$\displaystyle I=x[\ln(x)]^n-n\int [\ln(x)]^{n-1}dx$