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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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$\log x$ is usually used to refer to $\ln x$. Are you asking for help in evaluating the integral? –  Sanath Jun 8 at 0:23
    
Only in the case where logx does not refer to lnx. –  user155812 Jun 8 at 0:24
    
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 $u=x\log x$. –  André Nicolas Jun 8 at 0:26
    
Okay, that's good to know. Thanks for the clarification! –  user155812 Jun 8 at 0:29
    
I am reasonably sure that for all other notions of $\log$ there is no elementary antiderivative. –  André Nicolas Jun 8 at 0:36

2 Answers 2

up vote 5 down vote accepted

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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