Answer
$4 \ln (2) -\dfrac{3}{4}$
Work Step by Step
The area is given by $A=\int_1^{2} (x+1) \ln x \ dx$
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 $dv=(x+1) dx \implies v=\dfrac{dx}{x}$
$\displaystyle \int_1^{2} (x+1) \ln x \ dx=\ln x (\dfrac{x^2}{2}+x)-\int_1^{2} (\dfrac{x^2}{2}+x) (\dfrac{1}{x}) \ dx$
or, $=[\ln x (\dfrac{x^2}{2}+x)-\dfrac{x^2}{4}-x]_1^{2}$
or, $=[\ln (2) (\dfrac{2^2}{2}+2)-\dfrac{2^2}{4}-2]-[\ln (1) (\dfrac{1^2}{2}+1)-\dfrac{1^2}{4}-1]$
or, $=[4 \ln (2) -3]+\dfrac{9}{4}$
Therefore, the required area is: $Area=4 \ln (2) -\dfrac{3}{4}$
