(after having spent some time working in the wrong region):
The statement of the boundaries is ambiguous, because they apply equally well to the blue or the green region shown in the graph above. But it wasn't a complete waste of time to integrate in the wrong place, as I will describe below.
The equation of the circle is $ \ x^2 \ + (y-1)^2 \ = \ 1 \ $ , so the functions for the "upper and lower" semicircles are $ \ y \ = \ 1 \ \pm \ \sqrt{1-x^2} \ . $ The blue region is bounded above by the upper semi-circle and below by the line $ \ x + y = 2 \ $ . The intersection points of the line and circle are $ \ (0,2) \ \text{and} \ (1,1) \ . $ So the double integral is
$$ \int_0^1 \int^{1 + \sqrt{1-x^2}}_{2-x} \ x \ \ dy \ dx \ = \ \int_0^1 \ xy \ \vert^{1 + \sqrt{1-x^2}}_{2-x} \ \ dx $$
$$ = \ \int_0^1 \ x \ \left[ \ (1 + \sqrt{1-x^2}) \ - \ (2-x) \ \right] \ \ dx \ = \ \int_0^1 \ x^2 \ - \ x \ + \ x \sqrt{1-x^2} \ \ dx $$
$$ = \ \left[ \ \frac{1}{3}x^3 \ - \ \frac{1}{2}x^2 \ - \ \frac{1}{3} (1-x^2)^{3/2} \ \right] \ \vert_0^1 $$
$$ = \ \left( \ \frac{1}{3} \ - \ \frac{1}{2} \ - \ 0 \ \right) \ - \left( \ 0 \ - \ 0 \ - \ \frac{1}{3} \ \right) \ = \ \frac{1}{6} \ , $$
which apparently does not agree with the "given answer". So who's right?
Let's take the integration over the green region, which is bounded below entirely by the lower semi-circle, bounded above by the upper semi-circle over $ \ [-1 , 0 ] \ $ , and above by the line over $ \ [0 , 1] \ . $ So we have two integrals here:
$$ \int_{-1}^0 \int^{1 + \sqrt{1-x^2}}_{1 - \sqrt{1-x^2}} \ x \ \ dy \ dx \ \ + \ \ \int_0^1 \int^{2-x}_{1 - \sqrt{1-x^2}} \ \ x \ \ dy \ dx $$
$$ = \ \ \int_{-1}^0 \ \ x \ \left[ \ (1 + \sqrt{1-x^2}) \ - \ ({1 - \sqrt{1-x^2}}) \ \right] \ \ dx $$
$$ + \ \ \int_0^1 \ x \ \left[ \ (2-x) \ - \ (1 - \sqrt{1-x^2}) \ \right] \ dx $$
$$ = \ \ 2 \ \int_{-1}^0 \ \ x \ \sqrt{1-x^2} \ \ dx \ + \ \ \int_0^1 \ x \ - \ x^2 \ + \ x \sqrt{1-x^2} \ \ dx $$
$$ = \ \left[ \ - \ \frac{2}{3} (1-x^2)^{3/2} \ \right] \ \vert_{-1}^0 \ + \ \ \left[ \ \frac{1}{2}x^2 \ - \ \frac{1}{3}x^3 \ - \ \frac{1}{3} (1-x^2)^{3/2} \ \right] \ \vert_0^1 $$
$$ = \ \left[ \ -\frac{2}{3} \ + \ 0 \ \right] \ + \ \left[ \ \left( \ \frac{1}{2} \ - \ \frac{1}{3} \ - \ 0 \ \right) \ - \left( \ 0 \ - \ 0 \ - \ \frac{1}{3} \ \right) \ \right] $$
$$ = \ -\frac{2}{3} \ + \ \frac{1}{6} \ + \ \frac{1}{3} \ = \ -\frac{1}{6} \ . $$
What was the point of doing this? Together, the green and blue regions cover the entire circle, which is symmetrical about the $ \ y- $ axis. We are integrating $ \ x \ $ , a function with odd symmetry (about the $ \ y-$ axis) over this region. So the integral of this function over the circle is zero. And indeed, the integrals over the blue and green regions have opposite signs.
So I believe the result of $ \ \frac{1}{6} \ $ for the blue region is reliable. My suspicion is that the solver may have made an error summing the fractions (I've seen it happen...).
1.) Transform differentials:
$$\begin{gathered}
(x + y){e^{x + y}}\frac{1}{{{x^2}}}dydx \hfill \\
u = \frac{y}{x},v = x + y \hfill \\
du = - \frac{y}{{{x^2}}}dx + \frac{1}{x}dy \hfill \\
dv = dx + dy \hfill \\
du \wedge dv = - \frac{y}{{{x^2}}}dx \wedge dy + \frac{1}{x}dy \wedge dx \hfill \\
du \wedge dv = - \frac{{x + y}}{{{x^2}}}dx \wedge dy = \frac{{x + y}}{{{x^2}}}dy \wedge dx \hfill \\
\frac{1}{v}du \wedge dv = \frac{1}{{{x^2}}}dy \wedge dx \hfill \\
\frac{1}{{{x^2}}}dydx = \frac{1}{v}dudv \hfill \\
(x + y){e^{x + y}}\frac{1}{{{x^2}}}dydx = {e^v}dudv \hfill \\
\end{gathered}$$
2.) Find inverse function:
$$\begin{gathered}
u = \frac{y}{x},v = x + y \Rightarrow uv - y = x\frac{{{y^2}}}{{{x^2}}} = x \cdot {u^2} \hfill \\
\Rightarrow uv - x - y = x \cdot {u^2} - x \hfill \\
\Rightarrow uv - v = x \cdot ({u^2} - 1) \hfill \\
\Rightarrow x = \frac{{v(u - 1)}}{{{u^2} - 1}} = \frac{v}{{u + 1}} \hfill \\
v = x + y \Rightarrow \hfill \\
y = v - x = v - \frac{v}{{u + 1}} = v(1 - \frac{1}{{u + 1}}) = v \cdot \frac{u}{{u + 1}} \hfill \\
x = v \cdot \frac{1}{{u + 1}} \hfill \\
y = v \cdot u\frac{1}{{u + 1}} \hfill \\
\end{gathered} $$
3.) Transform boundaries:
$$\begin{gathered}
x = 0 \Rightarrow v = 0 \hfill \\
x = 1 \Rightarrow v = u + 1 \hfill \\
y = 0 \Rightarrow v \cdot u = 0 \Rightarrow u = 0 \vee v = 0 \hfill \\
y = x \Rightarrow u = 1 \hfill \\
\end{gathered}$$
4.) Integrate:
$$\begin{gathered}
\int\limits_0^1 {\int\limits_0^x {(x + y){e^{x + y}}\frac{1}{{{x^2}}}dydx = \int\limits_0^1 {\int\limits_0^{u + 1} {{e^v}dvdu} } } } = \int\limits_0^1 {({e^{u + 1}} - 1)du} \hfill \\
\int\limits_0^1 {({e^{u + 1}} - 1)du} = {e^2} - e - 1 \hfill \\
\end{gathered}$$
Best Answer
The Jacobian :
You have $x=u+v$ and $y=u-v$ then : $|J|=2$