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I have the derivative of a function $f(x)$: $f'(x) = \log(x)$, where $\log(x)$ is the natural logarithm. What's the original function $f(x)$ and what is that calculation called in English?

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    $\begingroup$ This is called "integrating $\log(x)$". $\endgroup$ Mar 26, 2013 at 14:19

4 Answers 4

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Hint : integrate by parts. $du = dx$, $v=\log{x}$.

$$\int \log{x} \, dx = x \log{x} - \int x \frac{d}{dx} (\log{x}) = x \log{x} - x + C$$

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  • $\begingroup$ Ah, integration it's called. But I'm a newbie, so I cannot follow the rest of your answer. Can you explain? $\endgroup$
    – user63495
    Mar 26, 2013 at 14:18
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    $\begingroup$ Integration by parts, there are some examples $\endgroup$
    – Blex
    Mar 26, 2013 at 14:52
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$[x \ln x -x\ ] + C$ is what you're looking for, it's called the anti-derivative of $\log x$

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    $\begingroup$ ...plus a constant....:) $\endgroup$
    – DonAntonio
    Mar 26, 2013 at 14:43
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    $\begingroup$ obviously... =) $\endgroup$
    – Bob
    Mar 26, 2013 at 15:33
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$x \log x - x$ try differentiating this.

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Here's a way to think about it if you haven't been introduced to integration.

$ \begin{aligned} \displaystyle \log{x} = (\log{x}+1)-1 = (x)'\log{x}+x(\log{x})'-1+0 = (x\log{x})'-(x)'+(\mathcal{C})' = (x\log{x}-x+\mathcal{C})' \end{aligned} $

Where $\mathcal{C}$ is any constant. Therefore your function is $f(x) = x\log{x}-x+\mathcal{C}$.

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