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The integral $$\int x^x \ln x\, dx= ?$$ I know of the integral $\int x^x dx$ can be further simplified as $\int e^{x\ln x} dx$. And this requires identity to simplify. What about the product in the integral $\int x^x\ln x\,dx=\int e^{x\ln x}\ln x\, dx.$ Is there any identity to be used for this one.

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2 Answers 2

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$$\int x^x (\ln x +1-1) dx= \int e^{x\ln x}(\ln x+1)dx -\int x^x dx$$ $$=\int e^{x\ln x} (x\ln x)'dx -\int x^x dx = e^{x\ln x}-\int x^x dx=x^x -\int x^x dx$$ There is now way to solve the last integral.

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    $\begingroup$ Yeah. That's very good approach. Now I can deal with the $\int x^xdx.$ This, I know how to deal with it. $\endgroup$
    – unknown
    Jan 30, 2019 at 21:27
  • $\begingroup$ You want power series? $\endgroup$
    – user625055
    Jan 30, 2019 at 21:28
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    $\begingroup$ Yes fort the unresolved integral above. I have seen it somewhere how to do it, but I have to think deep again. Thanks for your help. $\endgroup$
    – unknown
    Jan 30, 2019 at 21:30
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Define $g(x) = x^x$. Then $\ln g(x) = x\ln x$ and differentiating both sides $$\frac{g'(x)}{g(x)}=\ln x+1,$$ which means $g'(x) = x^x(\ln x + 1)$. Now, up to a constant $$x^x = \int g'(x)\,dx = \int x^x \ln x \,dx + \int x^x\,dx$$ thus $$\int x^x \ln x\,dx=x^x-\int x^x\,dx. $$

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    $\begingroup$ Or you can find $g'(x)$ directly: $(x^x)'=(\exp(x\ln x))'=\exp(x\ln x)\times(x \ln x)'=\exp(x\ln x)(1\times \ln x + \frac1x \times x)=\exp(x\ln x)(\ln x + 1))=x^x(\ln x + 1)$ $\endgroup$
    – trolley813
    Jan 31, 2019 at 6:04

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