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Find $$\int_{1}^{e}\frac{1}{\ln(x^x e^x)} dx.$$ Any hints please? Couldn't think of any approach. Even wolfram alpha fails to give an answer.

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    $\begingroup$ Actually, W|A doesn't give you an answer for the indefinite integral, but it computes the value for the definite one. Using logarithms laws and the fact $x>0$ $$\ln(x^x e^x)=\ln(x^x)+\ln(e^x)=x\ln(x)+x$$ and this is something you should know (and W|A knows) how to integrate. $\endgroup$
    – Galc127
    Sep 2, 2016 at 7:50
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    $\begingroup$ W&A yields $\log(2)$ with $\texttt{Integrate[1/Log[x^x E^x], {x, 1, E}]}$. wolframalpha.com/input/… $\endgroup$ Sep 3, 2016 at 2:01

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We have $$ \ln(x^xe^x)=\ln(x^x)+\ln(e^x)=x\ln x+x $$ and $$ D(1+\ln x)=\frac1x, $$ and $$ \frac{1}{x\ln x+x}=\frac1x\,\frac1{1+\ln x}, $$ so...

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  • $\begingroup$ Great.I wonder why wolfram could'nt answer it :/. $\endgroup$
    – user220382
    Sep 2, 2016 at 7:50
  • $\begingroup$ @SanchayanDutta: At least Mathematica is (or at least the older versions were) occasionally excessively worried about thing such as multivaluedness of complex powers. So may be WA won't do the simplification in the first line, because some of the rules need $x$ to be a positive real number. I would be surprised, if it cannot integrate it after that simplification. $\endgroup$ Sep 2, 2016 at 7:53
  • $\begingroup$ Having done that, the indefinite integral is $\ln(1+\ln x)$ and the definite integral is $\ln2$. $\endgroup$ Sep 2, 2016 at 7:58
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    $\begingroup$ W&A yields $\log(2)$ with $\texttt{Integrate[1/Log[x^x E^x], {x, 1, E}]}$. wolframalpha.com/input/… $\endgroup$ Sep 3, 2016 at 1:53

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