# Why do we ignore constant produced by integration of derivative when we derive integration by parts formula?

My question is somehow related to this question. What I would like to know is why we ignore the fact that integral of derivative of f(x) is equal to f(x) only up to a constant.

$$\int (\frac{d}{dx}[f(x)]) dx = f(x) + c$$

This has been very nicely answered in this question.

Derivation goes as follows in Pauls notes. Step 1. Start with the product rule.

$$\{uv\}' = uv' + vu'$$

Step 2. Integrate both sides.

$$\int \{uv\}' = \int udv + \int vdu$$

Step 3. Due to Fundamental Theorem of Calculus, part I, left hand side should simplify.

$$uv = \int udv + \int vdu$$

Step 4. From both sides subtract $\int vdu$

$$\int udv = uv-\int vdu ?$$

Now, what makes me confused is Step 3. It should simplify but it should also produce a constant. We could argue, that this constant will simplify with constants produced on the right hand side simply because of equivalency starting in Step 1, but my issue here is feels we are ignoring this constant and we don't keep track of them. This seems not so important on paper or on blackboard, but if we use a computer algebra system such as Sympy it will not recognise equation in Step 4 as true.

• It is implicit that $\int u\ dv$ has an integration constant hiding inside of it. – Doug M Jun 21 '18 at 1:29
• By the way, it is worth mentioning you are a little bit wrong. If $f_1'(x)=f_2'(x)=g(x) \forall x\in(a,b)$, then $f_1(x)=f_2(x)+C\forall x\in(a,b)$ for some $C\in\mathbb R$. The open interval is important. The method people due with indefinite integral, is somewhat informal, but I believe if one understand it deeply, proofs will be rigorous even if there is a subtle notation problem. If you want to make it clear, I would recommend to you that treating indefinite integral as a set (of antiderivative) may makes things crystal clear. – Tony Ma Jun 21 '18 at 11:26
• @TonyMa, that is interesting, could you maybe link a derivation that is following this approach? I would like to look it up. – Marek Jun 22 '18 at 15:12
• Well... I haven't found anyone treating antiderivative as a set (maybe I just can't find), but I think it is a good approach (of course it is okay to use the traditional approach if you understand the meaning of constant of integration). To briefly introduce my approach, I will let $$\int\frac{1}{x}dx=\{f:\frac{d}{dx}f=\frac{1}{x}\}=\{f:C_1,C_2\in\mathbb R, \text{dom}(f)=\mathbb R^-\cup\mathbb R^+, f(x)=\ln x+C_1\forall x\in\mathbb R^+,f(x)=\ln x+C_2\forall x\in\mathbb R^-\}$$ – Tony Ma Jun 23 '18 at 2:50
• By the way, I think all but a few people will have ever forgot the condition of open interval $(a,b)$, including me. However, I had a better understanding of antiderivative after I treated it as a set. – Tony Ma Jun 23 '18 at 2:56

Indeed, it is so that $\int(uv)' = uv + c$ where $c$ is an arbitrary constant.
So by step three we would have: $uv= (\int u'v) + (\int uv') - c$
But wait!   We intend that $\int u'v$ and $\int uv'$ will both eventually be evaluated as "something plus an arbitrary constant," and the sum of arbitrary constants is an arbitrary constant.   So by convenience, we can pull any arbitrary constants into them until we are ready.   That is okay if we ensure an arbitrary constant 'pops out' when we have evalated all indefinite integrals.
The convention is that if there's an indefinite integral written in an expression, it incorporates an arbitrary constant. Multiple instances of the same indefinite integral may have different constants implicit, however: recall the famous "proof" $$\int \frac{dx}{x} = \frac{x}{x} + \int \frac{x}{x^2} \, dx = 1 + \int \frac{dx}{x}.$$