Why can't $\int x(x-3)^8\ dx$ be integrated by parts? I've been given the task of integrating the following equation $$\int x(x-3)^8dx$$ but they've all been saying that it must be done via substitution and is impossible to do view integration by parts. I'm wondering why that is.
Video of solution by substition:
https://youtu.be/CXgsorgesS0?t=109
Attempt to solve it by parts: 
$$\int x(x-3)^8dx$$
$$ v=x,u'= (x-3)^8 $$
$$ u'=1, u = 1/9(x-3)^9$$
$$\int x(x-3)^8dx = (1/9)x(x-3)^9 -\int 1/9(x-3)^9$$
$$=(1/9)x(x-3)^9 -(1/90)(x-3)^{10}+C$$
 A: Note that the substitution $x \mapsto x \! + \! 3$ moves the $8^\text{th}$ power from the binomial onto the monomial $x$, making the integrand much easier to attack directly:
$$\int x(x-3)^8 \ \text{d}x \ = \ \int (x+3)x^8 \ \text{d}x \ = \ \int x^9 + 3x^8 \ \text{d}x$$
Relative to this, IBP involves more work.  Despite inefficient choices for $u$ and $\text{d}v$ relative to Parcly Taxel's answer (+1), the following demonstrates that IBP is not only possible, but can be carried out in more than one way.

$\textbf{Warning: }$Unnecessarily hard, but posting for pedagogical value.  If one were asked on a test to solve this integral by parts, we advise—in no uncertain terms—to use Parcly Taxel's choices for $u$ and $\text{d}v$.  For illustrative purposes only!

Letting $u = (x-3)^8$ and $\text{d}v = x$, we get:
$$\int x(x-3)^8 \ \text{d}x \ = \ \frac{1}{2}x^2(x-3)^8 - \int4x^2(x-3)^7 \ \text{d}x$$
To solve the newly-created integral, you can proceed similarly with $u = (x-3)^7$ and $\text{d}v = 4x^2$.  Continue as such until the binomial disappears with $\text{d}u = \text{d}x$; at this point the final integral of the form $\displaystyle \int v \ \text{d}u$ should look something like $\displaystyle \int cx^8 \ \text{d}x$ for some $c \in \mathbb{R}$.  Solve this integral and, lastly, add all of the terms together.
A: Yes, it can be done by parts, treating $(x-3)^8$ as the expression to integrate and $x$ as the expression to differentiate:
$$\int x(x-3)^8\,dx=x\cdot\frac19(x-3)^9-\int\frac19(x-3)^9\,dx=\frac19x(x-3)^9-\frac1{90}(x-3)^{10}+K=\frac1{30}(3x+1)(x-3)^9+K$$
To get to the book answer:
$$\frac1{30}(3x+1)(x-3)^9+K=\frac{3x-9+10}{30}(x-3)^9+K=\left(\frac{x-3}{10}+\frac13\right)(x-3)^9+K=\frac{(x-3)^{10}}{10}+\frac{(x-3)^9}3+K$$
A: Choose $x$ as the function to differentiate and $(x-3)^8$ as the function to integrate, then $$\int{x(x-3)^8}=\frac{x(x-3)^9}{9}-\frac{1}{9}\int{(x-3)^9}$$
$$=\frac{x(x-3)^9}{9}-\frac{(x-3)^{10}}{90}$$
