1. Interpret the integral

We need to evaluate

$$I=\int_1^e \left[\left(\frac{x}{e}\right)^{2x}-\left(\frac{e}{x}\right)^x\right]\ln x\,dx.$$

Rewrite the powers in exponential form:

$$\left(\frac{x}{e}\right)^{2x}=e^{2x(\ln x-1)}, \qquad \left(\frac{e}{x}\right)^x=e^{x(1-\ln x)}.$$

But a more useful simplification is:

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

2. Use the substitution $x=e^t$

Let

$$x=e^t \quad \Rightarrow \quad dx=e^t dt, \quad \ln x=t.$$

When $x=1$, $t=0$; when $x=e$, $t=1$.

Now simplify each term:

$$\left(\frac{x}{e}\right)^{2x}=\left(\frac{e^t}{e}\right)^{2e^t}=\left(e^{t-1}\right)^{2e^t}=e^{2e^t(t-1)}$$

and

$$\left(\frac{e}{x}\right)^x=\left(\frac{e}{e^t}\right)^{e^t}=\left(e^{1-t}\right)^{e^t}=e^{e^t(1-t)}.$$

So

$$I=\int_0^1 \left[e^{2e^t(t-1)}-e^{e^t(1-t)}\right] t e^t\,dt.$$

This form is not immediately convenient, so we instead look for derivatives of the given expressions in terms of $x$.

3. Differentiate the key functions

Let

$$f(x)=\left(\frac{x}{e}\right)^{2x}, \qquad g(x)=\left(\frac{e}{x}\right)^x.$$

Then

$$\ln f(x)=2x(\ln x-1).$$

Differentiating,

$$\frac{f'(x)}{f(x)}=2\ln x.$$

Hence

$$f'(x)=2\ln x\left(\frac{x}{e}\right)^{2x}.$$

So

$$\left(\frac{x}{e}\right)^{2x}\ln x=\frac12 f'(x).$$

Similarly,

$$\ln g(x)=x(1-\ln x).$$

Differentiating,

$$\frac{g'(x)}{g(x)}=-\ln x.$$

Thus

$$g'(x)=-\ln x\left(\frac{e}{x}\right)^x.$$

So

$$\left(\frac{e}{x}\right)^x\ln x=-g'(x).$$

4. Rewrite the integral using derivatives

Therefore,

$$I=\int_1^e \left[\frac12 f'(x)-\left(-g'(x)\right)\right]dx =\int_1^e \left(\frac12 f'(x)+g'(x)\right)dx.$$

So

$$I=\frac12\big[f(x)\big]_1^e+\big[g(x)\big]_1^e.$$

5. Evaluate the boundary values

For $f(x)=\left(\frac{x}{e}\right)^{2x}$:

$$f(e)=\left(\frac{e}{e}\right)^{2e}=1,$$ $$f(1)=\left(\frac{1}{e}\right)^2=\frac1{e^2}.$$

Hence

$$\frac12[f(x)]_1^e=\frac12\left(1-\frac1{e^2}\right)=\frac12-\frac1{2e^2}.$$

For $g(x)=\left(\frac{e}{x}\right)^x$:

$$g(e)=\left(\frac{e}{e}\right)^e=1,$$ $$g(1)=\frac{e}{1}=e.$$

Thus

$$[g(x)]_1^e=1-e.$$

6. Add the results

$$I=\left(\frac12-\frac1{2e^2}\right)+(1-e) =\frac32-e-\frac1{2e^2}.$$

7. Match with the options

This is exactly

$$\boxed{\frac32-e-\frac1{2e^2}}$$

which corresponds to Option B.

8. Compare with stored correct answer

Stored correct answer: B

Our derived answer: B

So they agree.