1. Rewrite the given curves in convenient form

The curves are:

• $y=\ln(x+e^2)$
• $x=\ln\left(\frac{2}{y}\right)$
• $x=\ln 2$
• and the region is above the line $y=1$.

From $x=\ln\left(\frac{2}{y}\right)$ we get $e^x=\frac{2}{y}\quad\Rightarrow\quad y=2e^{-x}.$

So the relevant curves are:

• $y=\ln(x+e^2)$
• $y=2e^{-x}$
• $x=\ln 2$
• $y=1$

2. Find the intersection points relevant to the enclosed region

(i) Intersection of $y=2e^{-x}$ with $y=1$

$2e^{-x}=1 \Rightarrow e^{-x}=\frac12 \Rightarrow x=\ln 2.$ So $(\ln 2,1)$ is one corner.

(ii) Intersection of $y=\ln(x+e^2)$ with $y=1$

$\ln(x+e^2)=1 \Rightarrow x+e^2=e \Rightarrow x=e-e^2.$ So $(e-e^2,1)$ is another point.

(iii) Intersection of $y=\ln(x+e^2)$ and $y=2e^{-x}$

We check at $x=0$: $\ln(0+e^2)=2,\qquad 2e^0=2.$ So they intersect at $(0,2)$.

Thus the enclosed region is bounded:

• below by $y=1$ from $x=e-e^2$ to $x=\ln 2$,
• on the left-upper side by $y=\ln(x+e^2)$ from $x=e-e^2$ to $x=0$,
• on the right-upper side by $y=2e^{-x}$ from $x=0$ to $x=\ln 2$.

3. Set up the area integral

The area above $y=1$ is $A=\int_{e-e^2}^{0}\big(\ln(x+e^2)-1\big)\,dx+\int_{0}^{\ln 2}\big(2e^{-x}-1\big)\,dx.$

4. Evaluate the first integral

Let $I_1=\int_{e-e^2}^{0}(\ln(x+e^2)-1)dx.$ Substitute $u=x+e^2 \Rightarrow du=dx.$ When $x=e-e^2$, $u=e$; when $x=0$, $u=e^2$. So $I_1=\int_e^{e^2}(\ln u-1)du.$

Now, $\int (\ln u-1)du = u\ln u-2u.$ Therefore, $I_1=\left[u\ln u-2u\right]_e^{e^2}.$

At $u=e^2$: $e^2\ln(e^2)-2e^2=2e^2-2e^2=0.$

At $u=e$: $e\ln e-2e=e-2e=-e.$

Hence, $I_1=0-(-e)=e.$

5. Evaluate the second integral

Let $I_2=\int_0^{\ln 2}(2e^{-x}-1)dx.$

We have $\int (2e^{-x}-1)dx=-2e^{-x}-x.$ So $I_2=\left[-2e^{-x}-x\right]_0^{\ln 2}.$

At $x=\ln 2$: $-2e^{-\ln 2}-\ln 2=-2\cdot \frac12-\ln 2=-1-\ln 2.$

At $x=0$: $-2e^0-0=-2.$

Thus, $I_2=(-1-\ln 2)-(-2)=1-\ln 2.$

6. Total area

$A=I_1+I_2=e+(1-\ln 2)=1+e-\ln 2.$

7. Match with the options

$\boxed{1+e-\ln 2}$ which is Option B.

8. Compare with stored correct answer

Stored correct answer: B

Our derived answer is also B, so they agree.