Ray10
  • Ray10
use integration by parts: \[\int\limits_{}^{} x \times \ln \left| x \right| dx\]
Mathematics
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schrodinger
  • schrodinger
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bahrom7893
  • bahrom7893
use u = ln|x|, dv = x
anonymous
  • anonymous
\[\begin{matrix}u=\ln|x|&dv=x~dx\\ du=\frac{\text{sgn}x}{|x|}&v=\frac{1}{2}x^2\end{matrix}\] where \[\text{sgn}x:=\begin{cases}-1&\text{for }x<0\\0&\text{for }x=0\\1&\text{for }x>0\end{cases}\]
anonymous
  • anonymous
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

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Ray10
  • Ray10
in this case, @bahrom7893 why does \[u = \ln \left| x \right|\] and x=dx?
anonymous
  • anonymous
$$\int\underbrace{\log|x|}_u\underbrace{x\,\mathrm{d}x}_{\mathrm{d}v}=\underbrace{\frac12 x^2}_v\underbrace{\log|x|}_u-\int \underbrace{\frac12x^2}_v\underbrace{\frac1x\,\mathrm{d}x}_{\mathrm{d}u}=\frac12x^2\log|x|-\frac12\int x\,\mathrm{d}x$$Can you take it from here?
Ray10
  • Ray10
@oldrin.bataku I can take it from there, but I'm wondering how u=ln x, in this case?
anonymous
  • anonymous
Recall that $$\frac{d}{dx}\log|x|=\frac1x$$ http://en.wikipedia.org/wiki/Natural_logarithm#The_natural_logarithm_in_integration
anonymous
  • anonymous
@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.
Ray10
  • Ray10
@SithsAndGiggles what is the sgn?
anonymous
  • anonymous
@Ray10, I wrote the definition in the same comment.
Ray10
  • Ray10
now I understand it more, thanks @SithsAndGiggles
anonymous
  • anonymous
@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 \(x\).
anonymous
  • anonymous
@SithsAndGiggles is correct but it could be simplified: $$du=\frac{\operatorname{sgn} x}{|x|}=\frac1x$$
Ray10
  • Ray10
so positive \[\frac{ 1 }{ x } = \frac{ sgn x}{ \left| x \right| }\] ?
anonymous
  • anonymous
forgot the \(dx\)
anonymous
  • anonymous
@Ray10 they're equal for both positive and negative \(x\); observe;$$\frac{\operatorname{sgn}3}{|3|}=\frac13\\\frac{\operatorname{sgn}(-2)}{|-2|}=\frac{-1}2=\frac1{-2}=-\frac12$$
Ray10
  • Ray10
Taught me something new! @oldrin.bataku appreciate it!
anonymous
  • anonymous
@Ray10 no problem! glad I could help
Ray10
  • Ray10
and @SithsAndGiggles ! Oh and one question, is there a thanking method on here? like I noticed the medal thing @oldrin.bataku
anonymous
  • anonymous
Generally you give medals to answers that helped you or that you otherwise liked. You can also leave feedback if you visit their profile.
Ray10
  • Ray10
thank you! :) @oldrin.bataku
anonymous
  • anonymous
No problem!

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