Let's say we have
$$\int(x+1)^2\ln(3x)\,dx$$
Using integration by parts, first we'd designate:
$u=\ln(3x)$
$du=\frac{1}{x}~\mathrm dx$
$v=\dfrac{(x+1)^3}{3}$
$dv=(x+1)^2~\mathrm dx$
From here I would have
$$\ln(3x) \cdot \frac{(x+1)^3}{3}\; - \;\frac{1}{3}\int\frac{(x+1)^3}{x}\,dx$$
From here it's clear that if I try to integrate the new integral I'll just get ever increasing powers of $(x+1)$.
I looked to an integration calculator online (I don't know if I'm allowed to link it but if you just replace the space with a hyphen and stick dot com at the end, that's it) which shows that
$v=\frac{x^3}{3} + x^2 + x$
If I expand out $(x+1)^3,\;$ I get $\;x^3+3x^2+3x+1$ and then taking that over 3 leaves $\frac{x^3}{3} + x^2 + x + \frac{1}{3}$
My assumption here is that we can drop the $\frac{1}{3}$ since we're also excluding the $c$ anyway. Is that assumption correct that we can drop the constant part since it's indistinguishable from the $c$? If that isn't what's going on then what am I missing?
With respect to your choice of $u$ (and $du$, which you calculated), and your choice of $v$, both work nicely.
And you were spot on using the standard form, having identified $u, du, v$: $$uv - \int v\, du$$
$$\int(x+1)^2\ln(3x)\,dx \;=\; \frac{(x+1)^3}{3}\ln(3x)\;-\;\frac 13\int\frac{(x+1)^3}{x}dx$$
But there is no need to get spooked about integrating the remaining integral. After all, it is simply a polynomial, divided by a monomial:
When we expand the numerator of the remaining integral, we get $$(x^3 + 3x^2 + 3x+1)$$ and when this is divided by $x$ (in denominator), we get the integral:
$$\begin{align} \frac 13 \int \frac{(x^3 + 3x^2 + 3x + 1)}{x}\; dx &= \frac 13 \int \left( x^2 + 3x + 3 + \frac 1x\right) dx \\ \\ &= \frac 13 \left(\int x^2 \, dx + \int 3x \,dx + \int 3 dx + \int \frac 1x\, dx\right)\\ \\ &=\frac 13\left(\frac{x^3}3 + \frac {3x^2}2 + 3x + \ln x\right) + c \\ \\ \end{align}$$
All together, that gives us $$\frac{(x+1)^3}{3}\ln(3x) \;-\; \frac 13\left(\frac{x^3}3 + \frac {3x^2}2 + 3x + \ln x\right) + c$$
$$ = \frac 13 \left((x+1)^3\ln(3x)\; - \;\frac {x^3}3 - \frac{3x^2}2 - 3x - \ln x\right) + c$$