Integration by parts of $x^5\ln(x)$ I have to integrate the function: $x^5\ln(x)$
My Attempt
$$\int(x^5\ln(x))dx$$
$u=\ln(x)$ and $du=\frac{1}{x}dx$
$dv=x^5dx$ and
$v=\frac{x^6}{6}$
Using this I can then integrate the function using the method
$$uv-\int v du$$
Which then substituting I get:
$$\frac{x^6\ln(x)}{6}-\int\frac{x^6}{6}\frac{1}{x}dx$$
$$\frac{x^6\ln(x)}{6}-\int\frac{x^5}{6}dx$$
$$\frac{x^6\ln(x)}{6}-\frac{x^6}{36}+C$$
My question is if I used a different u instead of the first u which I used my first attempt how will getting the same answer look like? I tried setting $u=x^5$ and it just sort of repeated itself when integrating, I know the answer is the same but how does one get there?
What I mean
$u=x^5$ and $du=5x^4$
$dv=\ln(x)dx$ and $v=x\ln(x)-x$
Then using the method previously stated I get:
$$x^6\ln(x)-x^6-\int(x\ln(x)-x)(5x^4)dx$$

how about
$u = x^3$
$du=3x^2 dx$
$\int(x^5\ln(x))dx = \int \frac{ux^2 \ln(u^\frac{1}{3})}{3x^2} du $
properties of logs to eliminate cube root
=$\frac{1}{3} \int \frac{u \ln(u)}{3} du $
=$ \frac{1}{9}\int u \ln(u) du $
= $ \frac{1}{9}(\frac{1}{2}u^2\ln(u) - \int \frac{1}{2}u^2 / udu )$
= $ \frac{1}{9}(\frac{1}{2}u^2\ln(u) - \int \frac{1}{2}u du) $
= $ \frac{1}{9}(\frac{1}{2}u^2\ln(u) - \frac{1}{4}u^2) $
=$ \frac{1}{18}x^6\ln(x^3) - \frac{1}{36}x^6 $
=$ \frac{1}{6}x^6\ln(x) - \frac{1}{36}x^6  $
$ \frac{1}{6}x^6\ln(x) - \frac{1}{36}x^6 +c $
