<s>[INST] 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: 
$$\int x(x-3)^8dx$$
$$ v=x,u'= (x-3)^8 $$
$$ u'=1, u = 1/9(x-3)^9$$
$$\int x(x-3)^8dx = (1/9)x(x-3)^9 -\int 1/9(x-3)^9$$
$$=(1/9)x(x-3)^9 -(1/90)(x-3)^{10}+C$$
 [/INST] 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:
$$\frac1{30}(3x+1)(x-3)^9+K=\frac{3x-9+10}{30}(x-3)^9+K=\left(\frac{x-3}{10}+\frac13\right)(x-3)^9+K=\frac{(x-3)^{10}}{10}+\frac{(x-3)^9}3+K$$
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