Answer
$$\ln 4$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = \frac{1}{{x + y}} \cr
& {\text{From the image of the rectangle shown below we obtain}} \cr
& {\text{the region }}R \cr
& R = \left\{ {\left. {\left( {x,y} \right)} \right|0 \leqslant x \leqslant 1,{\text{ }}0 \leqslant y \leqslant x} \right\} \cr
& {\text{The area }}A{\text{ of the region is:}} \cr
& A = \frac{1}{2}\left( 1 \right)\left( 1 \right) \cr
& A = \frac{1}{2} \cr
& {\text{The average Value of a Function Over a Region }}R{\text{ is}} \cr
& {\text{Average value}} = \frac{1}{A}\iint\limits_R {f\left( {x,y} \right)}dA,{\text{ then}} \cr
& {\text{Average value}} = \frac{1}{{1/2}}\int_0^1 {\int_0^x {\frac{1}{{x + y}}} dydx} \cr
& = 2\int_0^1 {\left[ {\int_0^x {\frac{1}{{x + y}}} dy} \right]dx} \cr
& {\text{Integrate with respect to }}y \cr
& = 2\int_0^1 {\left[ {\ln \left| {x + y} \right|} \right]_0^xdx} \cr
& = 2\int_0^1 {\left[ {\ln \left| {x + x} \right| - \ln \left| {x + 0} \right|} \right]dx} \cr
& = 2\int_0^1 {\left( {\ln 2x - \ln x} \right)dx} \cr
& = 2\int_0^1 {\ln \left( {\frac{{2x}}{x}} \right)} dx \cr
& = 2\ln 2\int_0^1 {dx} \cr
& {\text{Integrate}} \cr
& = 2\ln 2\left[ x \right]_0^1 \cr
& = 2\ln 2\left( {1 - 0} \right) \cr
& = \ln 4 \cr} $$