1. ## Reduction Forumula

I'm trying to prove:

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

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

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

This turn into a really complicated thing, and I can't see were I will be able to find $I$ 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:

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

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

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

This turn into a really complicated thing, and I can't see were I will be able to find $I$ 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

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

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