Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
7  
This is called "integrating $\log(x)$". –  Alex Becker Mar 26 '13 at 14:19

4 Answers 4

up vote 5 down vote accepted

$[x \ln x -x\ ] + C$ is what you're looking for, it's called the anti-derivative of $\log x$

share|improve this answer
5  
...plus a constant....:) –  DonAntonio Mar 26 '13 at 14:43
1  
obviously... =) –  Bob Mar 26 '13 at 15:33

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

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

share|improve this answer
    
Ah, integration it's called. But I'm a newbie, so I cannot follow the rest of your answer. Can you explain? –  Camil Staps Mar 26 '13 at 14:18
1  
Integration by parts, there are some examples –  Blex Mar 26 '13 at 14:52

$x \log x - x$ try differentiating this.

share|improve this answer

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 $f(x) = x\log{x}-x+\mathcal{C}$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.