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Does anyone know of a rule to tell which derivative to use when faced with an integral like this? (C is any constant, eg. 2400 or 4, etc).

$$\int \:\frac{C}{x}\mathrm{d}x$$

I know that $\ln(x) = \int\frac{1}{x}\mathrm{d}x$ so then $\int \:\frac{C}{x}\mathrm{d}x$ could equal $C\ln(x)$.

I also know that $\frac{1}{x}$ can be $x^{-1}$.

Both the $\ln(x)$ and $x^{-1}$ give different answers, so employing the wrong one is a problem. So how do we know which one to use and what situation?

An example, $\int \:\frac{2.6}{x}$. Which "derivatie method" (excuse my lack of proper terminology) would I use if I am trying to get the integral?

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    $\begingroup$ Both WHAT give different answers? $\endgroup$
    – user658409
    May 25, 2020 at 15:39
  • $\begingroup$ Sorry I am not great with the terminology to use. I'll try again to elaborate my question $\endgroup$
    – Anon
    May 25, 2020 at 15:43
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    $\begingroup$ $$ \int x^{-1}\, dx = \ln |x| + C $$ where $C$ is a piecewise constant, i.e. one constant when $x>0$ and a possibly different constant when $x<0.$ You seem to assert that SOMEHOW evaluating this integral gives you a different answer from that. HOW did you get that different answer, and what was it? $\qquad$ $\endgroup$
    – Michael Hardy
    May 25, 2020 at 15:50
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    $\begingroup$ "Which could be $\ln(x)$ or $x^{-2}$" The second one is incorrect. $\endgroup$
    – user658409
    May 25, 2020 at 15:58
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    $\begingroup$ @Anon : Note that $\dfrac d {dx} x^{-2} = -2x^{-3} .$ That is certainly nothing like $\ln x. \qquad$ $\endgroup$
    – Michael Hardy
    May 25, 2020 at 16:03

3 Answers 3

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What function, other than $\ln(x)$, would you differentiate to get $\frac{1}{x}$?

For instance $\frac{d}{dx} x^2 = 2x$, but $\frac{d}{dx} x^0 = \frac{d}{dx} 1 =0\neq x^{-1}$

I don't see how one could get a different answer than $C\ln(x)$

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  • $\begingroup$ Thanks Mike. Took me time to figure out what you meant (nothing wrong with how you said it). You would use the power rule if the x was raised to anything but 1 when it is the denominator of a fraction. $\endgroup$
    – Anon
    May 25, 2020 at 16:04
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If we try to integrate $x^{-1}$ using the power rule it replaces the function with a constant because $x^0=1$. Since it is a constant, its derivative is not actually $x^{-1}$ so we know it doesn't work. Since $-1$ is the only value that doesn't follow the power rule it's the only special case you have to remember.

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    $\begingroup$ But the power rule would yield $\dfrac{x^0} 0,$ which is quite different from $x^0. \qquad$ $\endgroup$
    – Michael Hardy
    May 25, 2020 at 16:04
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    $\begingroup$ @MichaelHardy I don't know whether I should argue that it's still constant or not given the absurdity of the expression. Constant-ish? $\endgroup$
    – CyclotomicField
    May 25, 2020 at 16:10
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    $\begingroup$ @CyclotomicField : I know of no literal sense in which it is constant. $\endgroup$
    – Michael Hardy
    May 25, 2020 at 16:12
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    $\begingroup$ @CyclotomicField : Besides, isn't the $0$ in the denominator precisely the occasion for noticing that the power rule doesn't hold in this case? $\endgroup$
    – Michael Hardy
    May 25, 2020 at 16:13
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    $\begingroup$ @MichaelHardy I certainly agree division by zero is an excellent reason for anything to fail, including the power rule and it avoids the claim that $x^0/0$ is constant. We can stereographically project the real line onto the circle and add the last point at infinity to reasonable interpret $1/0$ as a constant though so I'll leave it as is. $\endgroup$
    – CyclotomicField
    May 25, 2020 at 16:25
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The only way to get $\frac{1}{x}$ when dividing is by the use of $\ ln(x)$, because if we would try to do it by the power rule we would need to start with $x^0$ and when we derive that we get 0 since it would be something like $0 x^{-1}$ (even tough this is uncorrect), at the en $x^0$ is always a constant, therefore its derivative will always be 0. When integrating $\frac{\alpha}{x}$ we simply need to do the following:

$$\int{\frac{\alpha}{x}}=\alpha \int{\frac{1}{x}}=\alpha \ln\vert{{x}}\vert+C$$

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