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To perform an integration by parts, we compare an integration with $$\int u dv=uv-\int v du$$ trying to guess the best fitting for $u$ and $dv$, followed by derivative of $u$ to find $du$; and integration of $dv$ to find $v$. At that last step, we despise a constant which, at a regular infinitive integration process, it wouldn't be despised. My question is: why we do that? I know if we don't despise it, $uv$ can assume an undefined constant times variable value, but I'm looking for a conceptual justification.

adjan
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elcio.humphreys
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1 Answers1

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In truth, you could have it, but it would simply make everything more complicated.

$$\begin{align}\int u\ dv&=u(v+C)-\int v+C\ du\\&=uv+uC-\int v\ du-\color{purple}{\int C\ du}\\&=uv-\int v\ du+uC-\color{purple}{uC}\\&=uv-\int v\ du\end{align}$$

So it makes no difference in the end.

Simply Beautiful Art
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