I have used product rule to make $u=x$ and $v=e^{- \ln x}$. I get $u=1$ but I'm unsure what $dv$ should be. Would it be $dv=- \ln xe^{- \ln x}$ or $dv=- \ln e^{- \ln x}$ and then that would simplify to $dv= \ln x$?
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7$\begingroup$ Notice that $e^{-\ln x} = (e^{\ln x})^{-1} = x^{-1}$ $\endgroup$– user88595Jan 16, 2014 at 19:29
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$\begingroup$ As an exercise, you could also try integrating $x^{^\tfrac x{\ln x}}$ :-) $\endgroup$– LucianJan 17, 2014 at 9:25
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3 Answers
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If you remember your rules of logarithms, note that $$-\ln x=-1\cdot\ln x=\ln x^{-1},$$ so that $$e^{-\ln x}=x^{-1}=\frac1x.$$ Hence, $$xe^{-\ln x}=x\cdot\frac1x=1$$ for all $x>0,$ which makes our task a great deal simpler.
answered Jan 16, 2014 at 19:39
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$$e^{-\ln x} = \left( e^{\ln x}\right)^{-1} = (x)^{-1} = \frac 1x$$
$$x e^{-\ln x} = x\cdot \frac 1x$$Now, what is $x \cdot \dfrac 1x$ ?
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Full calculation:
$[xe^{-\ln x}]'=x'e^{-\ln(x)}+x[e^{-\ln(x)}]'=e^{-\ln(x)}-x[\ln(x)]'e^{-\ln(x)}=(1-x\cdot \frac 1x)e^{-\ln(x)}=0$
answered Jan 16, 2014 at 19:39