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$$\int (\log x+1)x^x\,dx$$

This integral was found from the MIT Integration Bee. After making several unsuccessful attempts, I decided to type it into Mathematica, only to find that Mathematica could only produce an answer for this integral in the case where $\log(x)$ referred to the natural logarithmic function $\ln(x)$.

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    $\begingroup$ $\log x$ is usually used to refer to $\ln x$. Are you asking for help in evaluating the integral? $\endgroup$ – user122283 Jun 8 '14 at 0:23
  • $\begingroup$ Only in the case where logx does not refer to lnx. $\endgroup$ – A is for Ambition Jun 8 '14 at 0:24
  • $\begingroup$ The term $\log$ in mathematics has the default meaning of natural logarithm. Widespread use of $\ln$ is relatively recent. The integration is easy, since $x^x=\exp(x\log x)$. Let $u=x\log x$. $\endgroup$ – André Nicolas Jun 8 '14 at 0:26
  • $\begingroup$ Okay, that's good to know. Thanks for the clarification! $\endgroup$ – A is for Ambition Jun 8 '14 at 0:29
  • $\begingroup$ I am reasonably sure that for all other notions of $\log$ there is no elementary antiderivative. $\endgroup$ – André Nicolas Jun 8 '14 at 0:36
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The form of the integrand suggests writing $$(1 + \log x)x^x = (1+\log x)e^{x \log x},$$ then observing that by the product rule, $$\frac{d}{dx}\bigl[x \log x\bigr] = x \cdot \frac{1}{x} + 1 \cdot \log x = 1 + \log x.$$ Consequently, the integrand is of the form $f'(x) e^{f(x)}$, and its antiderivative is simply $$e^{f(x)} = e^{x \log x} = x^x.$$

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Use the substitution $y=x^x$, then do logarithmic differentitation. to get $(1+\ln x)x^x=\frac{dy}{dx}$. Now you may go from here.

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