In the integration by parts equation $\int uv = u\int v - \int (u'\int v)$:
Consider $\int v = g(x)+c$. So the above equation becomes $\int uv = u\int v - \int u'.(g(x)+c)$. But in integration of say, $\ln x$ as shown below, this constant is neglected. Why?
$ \int\ln x.dx = \log x\int dx - \int(\frac{d}{dx}\log x.\int dx).dx $
$ \int\ln x.dx = \log x\int dx - \int (x\frac{1}{x}).dx$ (Step 2)
$ \int\ln x.dx = \log x\int dx - \int dx $
$ \int\ln x.dx = x\log x - x + C$
$ \int\ln x.dx = x(\log x -1) + C $
Why is it $x.\frac {1}{x}$ in step 2 and not $x.(\frac{1}{x} + C)$? Or will it make no difference either way?
Morally, adding a constant should not affect the result. We might try intuiting an reason (and you should regardless), but proof is a good way to settle these doubts.
Let $V'(x) = v(x)$ (choose an antiderivative of $v$). Then $$\begin{align}&\frac{d}{dx}\left[u(x)(V(x)+c) - \int u'(s)(V(s)+c)\,ds\right] =\\[2ex] &= u'(x)(V(x)+c)+u(x)v(x) -u'(x)(V(x)+c) \\[2ex] &=\fbox {$u(x)v(x)$}\end{align}$$
Where $c$ is an arbitrary real number. So no, choosing a different primitive/antiderivative of $v$ does not change the result of integration by parts, as we hoped.