intregration without substitution of $x^x \ln x$

Question

How do i integrate this without any substitution, purely algebraically :

$$x^x \ln ex$$

I've tried a lot but not have been able to:

$$x^x \ (ln x + 1) = \ln x^{x^x} + x^x$$

or $e^{x \ln x}\ln (x+1)$, i've tried all these methods

How do proceed after this?

Write using LaTeX, otherwise it's hard to understand what you mean. — Timbuc, Sep 30 '14 at 17:20
You're a member already six months and have posted 4 questions. I'd say it's about time you learn how. — Timbuc, Sep 30 '14 at 17:27
I know that this is the derivative of x^x, i want to know how to integrate above form to x^x — questioner, Sep 30 '14 at 17:45
$x^x \log(ex) = x^x(1 + \log x) = x^x (x\log x)' = x^x (\log x^x)' = (x^x)'$ — achille hui, Sep 30 '14 at 17:52
@user3397 $(\log f(x))' = \frac{f(x)'}{f(x)}$. So $x^x(\log x^x)' = x^x \frac{(x^x)'}{x^x} = (x^x)'$. — achille hui, Sep 30 '14 at 18:06

score 5 · Answer 1 · answered Sep 30 '14 at 17:49

5

Let $y=x^x$, then $$\ln y=x\ln x\implies \frac{y'}{y}=\ln x+1\\ y'=x^x(\ln x+1)=x^x\ln(ex)$$then $$\int x^x\ln(ex)=y+c=x^x+c$$

answered Sep 30 '14 at 17:49

Semsem

score 4 · Accepted Answer · answered Oct 04 '14 at 21:40

4

Comment turned into answer per request.

Notice $$x^x \log(ex) = x^x(1 + \log x) = x^x (x\log x)' = x^x (\log x^x)' = x^x \frac{(x^x)'}{x^x} = (x^x)'$$

We have

$$\int x^x \log(ex) dx = \int d(x^x) = x^x + \text{constant.}$$

answered Oct 04 '14 at 21:40

achille hui

2 Answers2