According to NTA, the required region $A_2$ which is shaded in crossed lines and comes out to be

$$ A_2=\int_1^2\left(\ln \frac{2}{y}-e^y+e^2\right) d y=1+e-\ln 2 $$

But according to us the required region $A_1$ comes out to be shaded in parallel lines, which can be obtained as

$$ \begin{aligned} A_1 &=\int_0^{\ln 2}\left(\ln \left(x+e^2\right)-2 e^{-x}\right) d x \\\\ &=\left.\left\{\left(x+e^2\right) \ln \left(x+e^2\right)-x+2 e^{-x}\right\}\right|_0 ^{\ln 2} \\\\ &=\left(\ln 2+e^2\right) \ln \left(\ln 2+e^2\right)-\ln 2+1 \\\\ & \quad-2 e^2-2 \\\\ &=\left(\ln 2+e^2\right) \ln \left(\ln 2+e^2\right)-\ln 2-2 e^2-1 \end{aligned} $$

Not given in any option.

The region asked in the question is bounded by three curves

$$ \begin{aligned} &y=\ln \left(x+e^2\right) \\\\ &x=\ln \left(\frac{2}{y}\right) \\\\ &x=\ln 2 \end{aligned} $$

There is only one region which satisfies above requirement and which also lies above line $y=1$

Line $y=1$ may or may not be the boundary of the region.