\begin{aligned} & \iint_D \frac{1-x^2 y^3}{\left(x+\sqrt{1-y^2}\right)^2} \mathrm{~d} x \mathrm{~d} y=\iint_D \frac{1}{\left(x+\sqrt{1-y^2}\right)^2} \mathrm{~d} x \mathrm{~d} y-\iint_D \frac{x^2 y^3}{\left(x+\sqrt{1-y^2}\right)^2} \mathrm{~d} x \mathrm{~d} y \\ =& 2 \iint_{D_1} \frac{1}{\left(x+\sqrt{1-y^2}\right)^2} \mathrm{~d} x \mathrm{~d} y-0=2 \int_0^{\frac{\sqrt{2}}{2}} \mathrm{~d} y \int_y^{\sqrt{1-y^2}} \frac{1}{\left(x+\sqrt{1-y^2}\right)^2} \mathrm{~d} x \\ =&-\left.2 \int_0^{\frac{\sqrt{2}}{2}} \frac{1}{x+\sqrt{1-y^2}}\right|_{x=y} ^{x=\sqrt{1-y^2}} \mathrm{~d} y=-\int_0^{\frac{\sqrt{2}}{2}} \frac{1}{\sqrt{1-y^2}} \mathrm{~d} y+2 \int_0^{\frac{\sqrt{2}}{2}} \frac{1}{y+\sqrt{1-y^2}} \mathrm{~d} y \end{aligned}
\begin{aligned} &=-\left.2 \int_0^{\frac{\sqrt{2}}{2}} \frac{1}{x+\sqrt{1-y^2}}\right|_{x=y} ^{x=\sqrt{1-y^2}} \mathrm{~d} y=-\int_0^{\frac{\sqrt{2}}{2}} \frac{1}{\sqrt{1-y^2}} \mathrm{~d} y+2 \int_0^{\frac{\sqrt{2}}{2}} \frac{1}{y+\sqrt{1-y^2}} \mathrm{~d} y \\ &=-\left.\arcsin y\right|_0 ^{\frac{\sqrt{2}}{2}}+2 \int_0^{\frac{\sqrt{2}}{2}} \frac{1}{y+\sqrt{1-y^2}} \mathrm{~d} y=-\frac{\pi}{4}+2 \int_0^{\frac{\sqrt{2}}{2}} \frac{1}{y+\sqrt{1-y^2}} \mathrm{~d} y \\ &y=\sin t \\ &=\frac{\pi}{4}+2 \int_0^{\frac{\pi}{4}} \frac{\cos t}{\sin t+\cos t} \mathrm{~d} t \end{aligned}

\begin{aligned} &=-\frac{\pi}{4}+\int_0^{\frac{\pi}{4}} \frac{(\sin t+\cos t)+(\cos t-\sin t)}{\sin t+\cos t} \mathrm{~d} t \\ &=\left.\ln (\sin t+\cos t)\right|_0 ^{\frac{\pi}{4}}=\frac{1}{2} \ln 2 . \end{aligned}
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