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

Question

Integrate:

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

This question looks like of the form of $$\int\ e^x(f(x)+f'(x))\,dx,$$ but don't know how to get the proper substitution?

Answer 1

First we have

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

Noting that

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

we conclude $$\int \left[\left(\frac{x}{e}\right)^x + \left(\frac{e}{x}\right)^x\right]\ln x\,dx=\left(\frac{x}{e}\right)^x-\left(\frac{e}{x}\right)^x+C.$$

Answer 2

Hint

Try with

\begin{align*} y & = \left(\frac{e}{x}\right)^{x}\\ \ln y & = x[1-\ln x]\\ \frac{y^{'}}{y} & = -\ln x. \end{align*}

Now for the first component think of a similar idea.

Answer 3

$\bf{My\; Solution::}$ Let $$\displaystyle I = \int \left[\left(\frac{x}{e}\right)^x+\left(\frac{e}{x}\right)^x\right]\cdot \ln (x) dx$$

Now we can write $$\displaystyle \left(\frac{x}{e}\right)^x = x^x \cdot e^{-x} = e^{x\cdot \left(\ln x-1\right)}.............(\star)\color{\red}\checkmark$$

And we can write $$\displaystyle \left(\frac{e}{x}\right)^x=x^{-x}\cdot e^x = e^{-x\cdot \left(\ln x-1\right)}.............(\star)\color{\red}\checkmark$$

Now We can write $$\displaystyle \left[\left(\frac{x}{e}\right)^x+\left(\frac{e}{x}\right)^x\right]=e^{x\cdot \left(\ln x-1\right)}+e^{-x\cdot \left(\ln x-1\right)}= 2\cos h(x\cdot \left[\ln x-1\right])$$

Bcz we can write $$\displaystyle \sin h y = \left(\frac{e^y-e^{-y}}{2}\right)$$ and $$\displaystyle \cos h y = \left(\frac{e^y+e^{-y}}{2}\right)$$

So Integral $$\displaystyle I = 2\int \cos h(x\cdot \left[\ln x-1\right])\cdot \ln(x)dx $$

Now Let $$x\cdot \left[\ln x-1\right]=u\;,$$ Then $$\ln(x)dx = du$$

So Integral $$\displaystyle I = 2\int \cos hudu = 2\sin hu+\mathcal{C} = e^{x\cdot \left(\ln x-1\right)}-e^{-x\cdot \left(\ln x-1\right)}+\mathcal{C}$$

So $$\displaystyle I = \int \left[\left(\frac{x}{e}\right)^x+\left(\frac{e}{x}\right)^x\right]\cdot \ln (x) dx = \left[\left(\frac{x}{e}\right)^x-\left(\frac{e}{x}\right)^x\right]+\mathcal{C}$$