How find this integral $I={\iint}_{D}\frac{(x+y)\ln(1+y/x)}{\sqrt{1-x-y}}\,dx\,dy$ find the value
$$I={\Large\iint}_{D}\dfrac{(x+y)\ln\left(1+\dfrac{y}{x}\right)}{\sqrt{1-x-y}}\,dx\,dy,$$
where $$D=\{(x,y)\mid x+y\le 1,x\ge 0,y\ge 0\}.$$

The integral is 
$$\int_0^1 \int_0^{1-x} \frac{ (x+y) \log(1 + \frac y x)}{ \sqrt{1 - x - y}} dy dx$$
It can be evaluated by parts.
$$
\begin{align*}
I_1 &= \int_0^{1-x} \frac{ (x+y) \log(1 + \frac y x)}{ \sqrt{1 - x - y}} dy \\  
  &=  \left[\log (1 + \frac yx)\frac{-2}{3}  \sqrt{-x-y+1} (x+y+2) \right ]_0^{1-x} \\ &  - \int_0^{1-x} \frac 1 {x+y}  \frac{-2}{3}  \sqrt{-x-y+1} (x+y+2) dy \\ 
 &= \frac 2 3 \int_0^{1-x} \frac{(x+y+2)\sqrt{1-x-y}}{x+y}dy \\ 
 &=  \frac 2 3 \left( -\frac{2}{3} \left(\sqrt{1-x} (x+5)-6 \sinh ^{-1}\left(\sqrt{\frac{1}{x}-1}\right)\right. \right )\\ 
\end{align*}$$
\begin{align*}
I_2 &= \int_0^1 \frac 2 3 \left( -\frac{2}{3} \left(\sqrt{1-x} (x+5)-6 \sinh ^{-1}\left(\sqrt{\frac{1}{x}-1}\right)\right. \right ) dx  \\ 
 &= - \frac 4 9  \left( \int_0^1  \sqrt{1-x} (x+5)dx  - 6 \int_0^1 \sinh^{-1} \sqrt{ \frac 1 x - 1 }dx\right )\\ 
 &= - \frac 4 9 \left( \left[ -\frac 2 5 (1 - x)^{3/2} (9 + x) \right ]_0^1  - 6 \int_0^1 \log \left( \sqrt{\frac  1 x  -1} +  \sqrt{ \left(\frac  1 x  -1 \right )  + 1}\right )dx \right )\\ 
 &= - \frac 4 9 \left( \frac {18}5 - 6 \left[ x \log \left(\sqrt{\frac{1}{x}-1}+\sqrt{\frac{1}{x}}\right)-\sqrt{1-x} \right ]_0^1 \right ) \\ 
 &= - \frac 4 9 \left( \frac{18}{5} - 6 \right ) = \frac {16}{15}
\end{align*}
