The area enclosed by the curves y = _ e (x + e ^ 2 ), x = _ e...

The area enclosed by the curves $y = \log_{e}$\left(x + {e}^{2}\right), x = \log_{e}$\left(\dfrac{2}{y}\right)$ and $x = \log_{e}2$, above the line $y = 1$ is

Now plotting the diagram of given functions we get,

According to JEE Mains, we have to calculate the required region ${A}_{2}$ which is shaded in crossed lines and comes out to be

${A}_{2} = {∫}_{1}^{2}\left(ln\dfrac{2}{y} - {e}^{y} + {e}^{2}\right)dy = 1 + e - \ln 2$

But according to the question the required region ${A}_{1}$ comes out to be shaded in parallel lines, which can be obtained as

${A}_{1} = {∫}_{0}^{\ln 2}\left(\ln \left(x + {e}^{2}\right) - 2{e}^{- x}\right)dx$

{= \left\{\left(x + {e}^{2}\right)\ln \left(x + {e}^{2}\right) - x + 2{e}^{- x}\right\}}_{0}^{\text\ln2}

$= \left(\ln 2 + {e}^{2}\right)\ln \left(\ln 2 + {e}^{2}\right) - \ln 2 + 1 - 2{e}^{2} - 2$

$= \left(\ln 2 + {e}^{2}\right)\ln \left(\ln 2 + {e}^{2}\right) - \ln 2 - 2{e}^{2} - 1$

Not given in any option.