How to evaluate $ \int\left[\left(\frac{x}{e}\right)^x+\left(\frac{e}{x}\right)^x\right]\ln(x)\,dx$?

Question

$$ \int \left[ \left( \frac {x}{e}\right) ^ x + \left( \frac {e}{x}\right) ^ x \right] \ln(x)\, dx$$

Here I have no clue from where to start with. I already tried some

• substitutions

• taking out ${e} ^ {x}$

But none seems to work please help

Answer 1

$$f(x)=\left(\frac{x}{e}\right)^x=\frac{x^x}{e^x}$$ $$f'(x)=\frac{e^xx^x(\ln(x)+1)-e^xx^x}{(e^x)^2}=\frac{e^xx^x\ln(x)}{e^{2x}}=\frac{x^x\ln(x)}{e^x}=\left(\frac{x}{e}\right)^x\ln(x)$$

$$g(x)=\left(\frac{e}{x}\right)^x=\left(\frac{e^x}{x^x}\right)$$ or $\left(\frac{x}{e}\right)^{-x}$ $$g'(x)=\frac{x^xe^x-e^xx^x(\ln(x)+1)}{x^{2x}}=\frac{-e^xx^x\ln(x)}{x^{2x}}=-\left(\frac{e}{x}\right)^x\ln(x)$$ our integral: $$\int h(x)dx=\int \left[f'(x)-g'(x)\right]dx=f(x)-g(x)+C=\left(\frac{x}{e}\right)^x-\left(\frac{e}{x}\right)^x+C$$

Answer 2

We have

$\frac{d}{dx}\left(x\ (\ln x - 1)\right) = (\ln x - 1) + 1 = \ln x$, and

$\left(\frac xe\right)^x = e^{x(\ln x - 1)}$

Answer 3

\begin{align*}\int \left[ \left( \frac {x}{e}\right) ^ x + \left( \frac {e}{x}\right) ^ x \right] \ln(x)dx\\ &=\bigg(\frac{x}{e}\bigg)-\bigg(\frac{e}{x}\bigg)^x+\mathcal{C}.\end{align*}