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 $$D=\{(x,y)\mid x+y\le 1,x\ge 0,y\ge 0\}.$$
 A: 
Squares and rectangles are "good" domains to integrate product functions.

Let $x=uv$ and $y=(1-u)v$ then $x+y=v$, $D$ corresponds to the $(u,v)$-domain $0\leqslant u,v\leqslant 1$ (a product domain), and the Jacobian of the change of variable $(u,v)\mapsto(x,y)$ is $v$ hence the function to be integrated becomes a product function, namely,
$$
I=\iint\frac{-v\log u}{\sqrt{1-v}}\cdot v\cdot \mathrm du\mathrm dv.
$$
This shows that $I=J\cdot K$
with
$$
J=\int_0^1(-\log u)\cdot \mathrm du,\qquad K=\int_0^1\frac{v^2}{\sqrt{1-v}}\cdot \mathrm dv.
$$
Note that
$J=\left.u-u\log u\right|_0^1=1$. Using the change of variable $v=1-w^2$ with Jacobian $2w$,
$$
K=\int_0^1\frac{(1-w^2)^2}{w}\cdot 2w\cdot \mathrm dw=2\left.\left(w-\tfrac23w^3+\tfrac15w^5\right)\right|_0^1=\tfrac{16}{15}.
$$
This yields
$$
I=1\cdot\tfrac{16}{15}.
$$
A: 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*}
A: $\displaystyle{%
I
=
{\Large\iint}_{\!\!D}
\!\!
{\left(x + y\right)\ln\left(1 + y/x\right)
 \over
 \sqrt{1 - x - y\,}}\,{\rm dx}\,{\rm d}y\,,\qquad
D \equiv
\left\lbrace
\left(x,y\right)\ \ni\ x \geq 0\,,\ y \geq 0\,,\ x + y \leq 1
\right\rbrace}$
\begin{align}
-&-------------------------------------\\
&I
=
\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}
{\left(x + y\right)\ln\left(1 + y/x\right)
 \over
 \sqrt{1 - x - y\,}}\,
\Theta\left(x\right)\,\Theta\left(y\right)\,\Theta\left(1 - x - y\right)
{\rm dx}\,{\rm d}y
\\[3mm]&=
\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}
{y\,\ln\left(y/x\right) \over \sqrt{1 - y\,}}\,
\Theta\left(x\right)\,\Theta\left(y - x\right)\,\Theta\left(1 - y\right)\,
{\rm dx}\,{\rm d}y
\\[3mm]&=
\int_{0}^{1}{\rm d}y\,
{y\,\ln\left(y\right) \over \sqrt{1 - y\,}}\int_{0}^{y}{\rm d}x\,
-
\int_{0}^{1}{\rm d}y\,
{y \over \sqrt{1 - y\,}}\int_{0}^{y}{\rm d}x\,\ln\left(x\right)
\\[3mm]&=
\int_{0}^{1}{\rm d}y\,
{y^{2}\,\ln\left(y\right) \over \sqrt{1 - y\,}}
-
\int_{0}^{1}{\rm d}y\,{y \over \sqrt{1 - y\,}}
\left[y\ln\left(y\right) - y\right]
=
\int_{0}^{1}{\rm d}y\,{y^{2} \over \sqrt{1 - y\,}}
\\[3mm]&=
\int_{0}^{1}\left[%
\left(1 - y\right)^{3/2} - 2\left(1 - y\right)^{1/2} + \left(1 - y\right)^{-1/2}
\right]\,{\rm d}y
\\[3mm]&=
\left.\vphantom{\Huge A}
-\,{2 \over 5}\left(1 - y\right)^{5/2}
+
{4 \over 3}\left(1 - y\right)^{3/2}
-
2\left(1 - y\right)^{1/2}
\right\vert_{0}^{1}
=
{2 \over 5} - { 4\over 3} + 2
=
{16 \over 15}
\end{align}
$$
\begin{array}{|c|}\hline\\
\color{#ff0000}{\large\quad%
I
=
{\Large\iint}_{\!\!D}
\!\!
{\left(x + y\right)\ln\left(1 + y/x\right)
 \over
 \sqrt{1 - x - y\,}}\,{\rm dx}\,{\rm d}y
\color{#000000}{\ =\ }
{16 \over 15}\quad}
\\ \\ \hline
\end{array}
$$
A: Here is how you advance

$$ I={\iint}_{D}\dfrac{(x+y)\ln\left(1+\dfrac{y}{x}\right)}{\sqrt{1-x-y}}\,dx\,dy={\int}_{0}^{1}\int_{0}^{1-y}\dfrac{(x+y)\ln\left(1+\dfrac{y}{x}\right)}{\sqrt{1-x-y}}\,dx\,dy. $$

Plot the region to see what's going on. 
A: my other solution:

$$I=\int\int_{D}\dfrac{(x+y)\ln{(1+\dfrac{y}{x})}}{\sqrt{1-x-y}}dxdy=\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{1}{\sin{t}+\cos{t}}}\dfrac{r(\sin{t}+\cos{t})\ln{(1+\tan{t})}}{\sqrt{1-r(\cos{t}+\sin{t}}}rdr$$

then
let $u=\sqrt{1-r(\cos{t}+\sin{t})}$

$$\Longrightarrow I=2\int_{0}^{\frac{\pi}{2}}\dfrac{\ln{(1+\tan{t})}}{(\cos{t}+\sin{t})^2}dt\cdot\int_{0}^{1}(1-2u^2+u^4)dy$$

then 

$$I=\dfrac{16}{15}\int_{0}^{\frac{\pi}{2}}\dfrac{\ln{(1+\tan{t})}}{(1+\tan{t})^2}d\tan{t}$$

so

$$I=-\dfrac{16}{15}\int_{0}^{\frac{\pi}{2}}\ln{(1+\tan{t})}d\dfrac{1}{1+\tan{t}}=\dfrac{16}{15}$$

Have other nice methods ? Thank you 
