The integral $$\int_1^e \left\{\left(\frac{x}{e}\right)^{2x} - \left(\frac{e}{x}\right)^x\right\} \log_e x \, dx$$ is equal to

We have to evaluate the definite integral

$$I=\int_{1}^{e}\Bigl\{\left(\tfrac{x}{e}\right)^{2x}-\left(\tfrac{e}{x}\right)^{x}\Bigr\}\,\log_e x\;dx.$$

It will be convenient to treat the two parts separately. First observe the simple but very useful differentiation facts.

For the function $$y_1(x)=\left(\tfrac{x}{e}\right)^{2x},$$ take natural logarithm on both sides:

$$\ln y_1 = 2x\bigl(\ln x-1\bigr).$$

Differentiating this with respect to $$x$$ gives

$$\frac{1}{y_1}\frac{dy_1}{dx}=2\ln x.$$

Therefore

$$\frac{dy_1}{dx}=2\ln x\;\left(\tfrac{x}{e}\right)^{2x}.$$

Re-writing, we get the integral representation

$$\left(\tfrac{x}{e}\right)^{2x}\ln x\;dx=\frac12\,d\!\left[\left(\tfrac{x}{e}\right)^{2x}\right].$$

Now consider the second function $$y_2(x)=\left(\tfrac{e}{x}\right)^{x}.$$ Its logarithm is

$$\ln y_2 = x\bigl(1-\ln x\bigr).$$

Differentiating again with respect to $$x$$ gives

$$\frac{1}{y_2}\frac{dy_2}{dx}=1-\ln x-\frac{x}{x}= -\ln x.$$

Hence

$$\frac{dy_2}{dx}=-\ln x\;\left(\tfrac{e}{x}\right)^{x},$$

and so

$$\left(\tfrac{e}{x}\right)^{x}\ln x\;dx = -\,d\!\left[\left(\tfrac{e}{x}\right)^{x}\right].$$

Armed with these two differential identities we can attack the given integral. Split it into two parts:

$$I=\int_{1}^{e}\left(\tfrac{x}{e}\right)^{2x}\ln x\;dx-\int_{1}^{e}\left(\tfrac{e}{x}\right)^{x}\ln x\;dx.$$

Using the first identity, the first integral becomes

$$\int_{1}^{e}\left(\tfrac{x}{e}\right)^{2x}\ln x\;dx=\frac12\int_{1}^{e}d\!\left[\left(\tfrac{x}{e}\right)^{2x}\right] =\frac12\left[\left(\tfrac{x}{e}\right)^{2x}\right]_{1}^{e}.$$

The second identity transforms the second integral as follows:

$$\int_{1}^{e}\left(\tfrac{e}{x}\right)^{x}\ln x\;dx = -\int_{1}^{e}d\!\left[\left(\tfrac{e}{x}\right)^{x}\right]= -\left[\left(\tfrac{e}{x}\right)^{x}\right]_{1}^{e}.$$

But in $$I$$ this entire integral appears with a minus sign, so

$$-\,\Bigl(\text{second integral}\Bigr)=+\left[\left(\tfrac{e}{x}\right)^{x}\right]_{1}^{e}.$$

Putting the two transformed pieces together we have

$$I=\frac12\left[\left(\tfrac{x}{e}\right)^{2x}\right]_{1}^{e}+\left[\left(\tfrac{e}{x}\right)^{x}\right]_{1}^{e}.$$

Now we only have to substitute the limits $$x=1$$ and $$x=e$$.

First bracket:

At $$x=e$$, $$\left(\tfrac{e}{e}\right)^{2e}=1^{2e}=1.$$

At $$x=1$$, $$\left(\tfrac{1}{e}\right)^{2}= \frac{1}{e^{2}}.$$

So

$$\frac12\Bigl(1-\frac1{e^{2}}\Bigr)=\frac12-\frac{1}{2e^{2}}.$$

Second bracket:

At $$x=e$$, $$\left(\tfrac{e}{e}\right)^{e}=1^{e}=1.$$

At $$x=1$$, $$\left(\tfrac{e}{1}\right)^{1}=e.$$

Thus

$$\left[\,\cdots\right]_{1}^{e}=1-e.$$

Combining these two evaluated expressions gives

$$I=\left(\frac12-\frac{1}{2e^{2}}\right)+(1-e)=\frac32-e-\frac{1}{2e^{2}}.$$

Hence, the correct answer is Option A.