Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

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 $\ln(x)$.

share|cite|improve this question
$\log x$ is usually used to refer to $\ln x$. Are you asking for help in evaluating the integral? – Sanath K. Devalapurkar Jun 8 '14 at 0:23
Only in the case where logx does not refer to lnx. – A is for Ambition Jun 8 '14 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 '14 at 0:26
Okay, that's good to know. Thanks for the clarification! – A is for Ambition Jun 8 '14 at 0:29
I am reasonably sure that for all other notions of $\log$ there is no elementary antiderivative. – André Nicolas Jun 8 '14 at 0:36
up vote 7 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 $$e^{f(x)} = e^{x \log x} = x^x.$$

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.