Should I add $C$ (constant of integration) before or after calculation (or does it matter)?

Question

For example, determine $\int \left(\frac{1}{2x+1}\right)dx$.

Given that $f(x)$ = $\ln(2x+1)$ and $f'(x)$ = $\ln\left(\frac{2}{2x+1}\right)$.

Would this be $\frac{1}{2} \int\left (\frac{2}{2x+1}\right)dx = \frac{1}{2} (\ln(2x+1) + C)$ or $\frac{1}{2} \int\left (\frac{2}{2x+1}\right)dx = \frac{1}{2} (\ln(2x+1)) + C$?

Answer 1

It doesn't matter. Since $c$ is arbitrary, these are the same answers.

Answer 2

A bit more waffle:

Integration is not enough to tell you the $y$-intercept of the resulting function, since differentiation threw that information away. The constant of integration is effectively simply telling you where the $y$-intercept of the antiderivative is.

In your case, you have got the choice of $\frac{1}{2}C$ or simply $C$ as your constants of integration; that is to say, you have the choice of $y$-value when $x=0$ of $\frac{1}{2}C$ in the first answer, or $C$ in the second answer.

But if you wanted to choose constants of integration so that the $y$-intercept is $N$, say, you can do it exactly the same in either case: just pick $C = 2N$ in the first case, or $C = N$ in the second case.

Since you can get exactly the same set of functions from varying $C$ in the first case as you can in the second, the two expressions are effectively the same; both answers yield the same set of antiderivatives.

Answer 3

I don't like the other answers, because the root of the problem is that you did not understand the meaning of the so-called 'integration constant'. $ \def\lfrac#1#2{{\large\frac{#1}{#2}}} $

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$\int f(x)\ dx$ denotes an anti-derivative of $f(x)$ with respect to $x$, namely an expression $g(x)$ that when differentiated with respect to $x$ gives $f(x)$. Symbolically you want to find $g(x)$ given that $\lfrac{d(g(x))}{dx} = f(x)$.

If you are given $f(x)$ and an anti-derivative $F(x)$ of $f(x)$, then you do have $\lfrac{d(F(x))}{dx} = f(x)$, but for any constant $c$ you also have $\lfrac{d(F(x)+c)}{dx} = \lfrac{d(F(x))}{dx} + \lfrac{dc}{dx}$ $= f(x) + 0 = f(x)$ by the basic properties of differentiation, and so you cannot determine $\int f(x)\ dx$ without having more information.

However, if $\lfrac{d(F(x))}{dx} = f(x)$ at every $x∈I$ where $I$ is some interval of the real line, then we can in fact conclude that $\int f(x)\ dx = F(x) + C$ for some constant $C$ (by something called Rolle's theorem). Take note, you do not know what constant $C$ is! Also, for any constant $k$, we have $k · \int f(x)\ dx = k · ( F(x) + C )$ $= k·F(x)+k·C$ for some constant $C$. Since we are always working with constants from a field $F$, and $\{ k·C : C∈F \} = F$, we also can deduce $k · \int f(x)\ dx = k·F(x)+D$ for some constant $D$.

As mentioned in a comment, you must actually know how the unknown constant 'propagates' to the final expression. For example, $\exp(\int f(x)\ dx) = \exp(F(x)+C)$ $= \exp(F(x))·\exp(C)$ for some constant $C$. If you are working in the reals, then $\exp(C) > 0$, so you can conclude that $\exp(\int f(x)\ dx) = \exp(F(x))·D$ for some constant $D > 0$. But if you are working in the complex numbers, then $\exp(C)$ can be any complex number except $0$, so you would instead conclude $\exp(\int f(x)\ dx) = \exp(F(x))·D$ for some constant $D ≠ 0$.