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:
$$x^6\ln(x)-x^6-\int(x\ln(x)-x)(5x^4)dx$$

Setting $u = x^{5}$, $v' = \ln x$ and noting that $$\int \ln x = x \ln x - x + C$$ yields
\begin{align}
I &= \int x^{5} \ln x dx \\
&= x^{6} \ln x - x^{6} - 5 \int (x^{5} \ln x - x^{5}) dx \\
&= x^{6} \ln x - x^{6} - 5 I + \frac{5 x^{6}}{6} + C \\
&= x^{6} \ln x - \frac{x^{6}}{6} - 5 I + C \\
\implies 6I &= x^{6} \ln x - \frac{x^{6}}{6} + C \\
\implies I &= \frac{ x^{6} \ln x }{6} - \frac{x^{6}}{36} + C
\end{align}
