Question

use integration by parts:
\[\int\limits_{}^{} x \times \ln \left| x \right| dx\]

Answer 1

use u = ln|x|, dv = x

Answer 2

\[\begin{matrix}u=\ln|x|&dv=x~dx\\
du=\frac{\text{sgn}x}{|x|}&v=\frac{1}{2}x^2\end{matrix}\]
where
\[\text{sgn}x:=\begin{cases}-1&\text{for }x<0\\0&\text{for }x=0\\1&\text{for }x>0\end{cases}\]

Answer 3

put lnx=t so x=e^t
now differentiate it we get dx=e^t dt
now putting it back in the integral we get\[\int\limits e ^{t}t dt\]
now integrate by parts

Answer 4

in this case, @bahrom7893 why does \[u = \ln \left| x \right|\] and x=dx?

Answer 5

$$\int\underbrace{\log|x|}_u\underbrace{x\,\mathrm{d}x}_{\mathrm{d}v}=\underbrace{\frac12 x^2}_v\underbrace{\log|x|}_u-\int \underbrace{\frac12x^2}_v\underbrace{\frac1x\,\mathrm{d}x}_{\mathrm{d}u}=\frac12x^2\log|x|-\frac12\int x\,\mathrm{d}x$$Can you take it from here?

Answer 6

@oldrin.bataku I can take it from there, but I'm wondering how u=ln x, in this case?

Answer 7

Recall that $$\frac{d}{dx}\log|x|=\frac1x$$
http://en.wikipedia.org/wiki/Natural_logarithm#The_natural_logarithm_in_integration

Answer 8

@Ray10 we can't integrate things with \(\log\) in them easily at all unless we can substitute them out. Here, our other option is to integrate something comparatively nice (a power of \(x\)) so we chose it instead.

Answer 9

@SithsAndGiggles what is the sgn?

Answer 10

@Ray10, I wrote the definition in the same comment.

Answer 11

now I understand it more, thanks @SithsAndGiggles

Answer 12

@Ray10 \(\operatorname{sgn} x\) is the sign of the number; it's \(\operatorname{sgn} 0=0\), \(\operatorname{sgn} x=-1\) for negative \(x\), \(\operatorname{sgn} x=1\) for positive \(x\).

Answer 13

@SithsAndGiggles is correct but it could be simplified: $$du=\frac{\operatorname{sgn} x}{|x|}=\frac1x$$

Answer 14

so positive \[\frac{ 1 }{ x } = \frac{ sgn x}{ \left| x \right| }\] ?

Answer 15

forgot the \(dx\)

Answer 16

@Ray10 they're equal for both positive and negative \(x\); observe;$$\frac{\operatorname{sgn}3}{|3|}=\frac13\\\frac{\operatorname{sgn}(-2)}{|-2|}=\frac{-1}2=\frac1{-2}=-\frac12$$

Answer 17

Taught me something new! @oldrin.bataku appreciate it!

Answer 18

@Ray10 no problem! glad I could help

Answer 19

and @SithsAndGiggles !
Oh and one question, is there a thanking method on here? like I noticed the medal thing @oldrin.bataku

Answer 20

Generally you give medals to answers that helped you or that you otherwise liked. You can also leave feedback if you visit their profile.

Answer 21

thank you! :) @oldrin.bataku

Answer 22

No problem!