Answer
$2+2 \ln (2)$
Work Step by Step
Consider $I= \iint_{R} f(x,y) dA$
or, $= \int_{1}^{2} \int_{-1/x}^{1/x} (x+1) dy dx$
or, $= \int_{1}^{2}[\dfrac{x^2}{2}+x]_{-1/x}^{1/x} dx$
or, $=\int_{1}^{2} 2+\dfrac{2}{x} dx$
Thus, we have $I=2x+2 \ln x]_1^2=2+2 \ln (2)$