Answer
$8 \ln (4)+e^{1/2}$
Work Step by Step
Here, we have: $y=\ln (x)$ and $y=1 -\ln x$
Suppose that $y=y \implies \ln x=1-\ln x$
or, $2 \ln x =1 \implies x=e^{1/2}$
Here, the area can be expressed as: $A=\int_{e^{1/2}}^{4} [\ln x-(1-\ln x)] \ dx$
or, $=\int_{e^{1/2}}^{4} (2\ln x-1) \ dx$
or, $=(2 x\ln x-2x-x)_{e^{1/2}}^4$
or, $=(2 x\ln x-3x)_{e^{1/2}}^4$
or, $=[2 (4)\ln (4)-3(4)]-[2 e^{1/2} \ln (e^{1/2}) -3(e^{1/2})]$
Therefore, the required area is: $Area=8 \ln (4)+e^{1/2}$
