Answer
$\dfrac{e^2+1}{2}$
Work Step by Step
The area is given by $A=\int_1^{e} x \ln x \ dx$
We will solve the given integral by using integrate-by-parts formula such as: $\int udv=uv-\int v du$
Here, $u=\ln x$ and $dv=x dx \implies v=\dfrac{x^2}{2}$
$\displaystyle \int_1^{e} x \ln x \ dx=\ln x (\dfrac{x^2}{2})-\int_1^{e} (\dfrac{x^2}{2}) (\dfrac{1}{x}) \ dx$
or, $=[\ln x (\dfrac{x^2}{2})-\dfrac{x^2}{4}]_1^{e}$
or, $=[\ln e (\dfrac{e^2}{2})-\dfrac{e^2}{4}]-[\ln 1 (\dfrac{1^2}{2})-\dfrac{1}{4}]$
or, $=\dfrac{e^2}{2}-\dfrac{e^2}{4}+\dfrac{1}{4}$
Therefore, the required area is: $Area=\dfrac{e^2+1}{2}$
