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:
$$x^6\ln(x)-x^6-\int(x\ln(x)-x)(5x^4)dx$$
 A: From what you have you can write the same as the following and proceed to get the original answer
$$I=x^6\ln(x)-x^6-\int(x\ln(x)-x)(5x^4)dx=x^6\ln(x)-x^6-\int5x^5dx-5I$$
A: \begin{align}
\int x^{5}\cdot \ln (x)\; {\rm d} x 
=&
\int \underbrace{ x^{5}}_{u'}\cdot \underbrace{\ln (x)}_{v}\; {\rm d} x 
\\
=&
\int \underbrace{ D_x\left(\frac{x^{6}}{6}\right)}_{u'}\cdot \underbrace{\ln (x)}_{v}\; {\rm d} x 
\\
=&
\underbrace{\left(\frac{x^{6}}{6}\right)}_{u}
\cdot 
\underbrace{\ln (x)}_{v}
-
\int \underbrace{\left(\frac{x^{6}}{6}\right)}_{u}\cdot 
\underbrace{D_x \ln (x)}_{v'}\; {\rm d} x 
\\
=&
\underbrace{\left(\frac{x^{6}}{6}\right)}_{u}
\cdot 
\underbrace{\ln (x)}_{v}
-
\int \underbrace{\left(\frac{x^{6}}{6}\right)}_{u}\cdot 
\underbrace{\frac{1}{x}}_{v'}\; {\rm d} x 
\\
=&
\frac{x^{6}\ln (x)}{6}
-
\int \frac{x^{5}}{6}\; {\rm d} x 
\\
=&
\frac{x^{6}\ln (x)}{6}
-
\frac{x^{6}}{36}+C
\\
\end{align}
A: As $\dfrac{d(x^m\ln x)}{dx}=x^{m-1}+mx^{n-1}\ln x$
$$x^m\ln x+C=\int x^{m-1}\ dx+m\int x^{n-1}\ln x\ dx$$
$$\iff m\int x^{n-1}\ln x\ dx=x^m\ln x-\dfrac{x^m}m+C$$
A: 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 $
A: Alternatively, one may start with
$$
\int x^tdx=\frac{x^{t+1}}{t+1},\qquad t\ne-1,
$$ then one may differentiate with respect to $t$ obtaining
$$
\int x^t\ln x\:dx=\frac{x^{t+1}\ln x}{t+1}-\frac{x^{t+1}}{(t+1)^2},\qquad t\ne-1,
$$ here $t:=5$ gives the desired result.
A: 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}
