Change of variables Double integral

$$ x+y = r(\cos\theta+\sin\theta) = r\sqrt{2}\sin\left(\theta+\frac\pi2\right). $$ So $$ \iint_A \frac{dx\,dy}{(x^2+y^2)^2} = \int_0^{\pi/2} \int_{(\csc(\theta+\pi/4))/\sqrt{2}}^1 \frac{r\,dr \, d\theta}{r^4}. $$ $$ = \int_0^{\pi/2} \left(\int_{(\csc(\theta+\pi/4))/\sqrt{2}}^1\ \frac{dr}{r^3} \right) \, d\theta $$

If you rotate your domain by $45$ degrees your line becomes $y=\sqrt{2} / 2$ and the integrand is invariant, so you get (also use the simmetry with respect to $x=0$): $$ \iint_A \frac{1}{(x^2+y^2)^2} = 2\int_0^{\sqrt 2/2}\,dx\int_{\sqrt 2/2}^{\sqrt{1-x^2}} \frac{dy}{(x^2+y^2)} $$

Here is another answer

$$\iint_A \frac{1}{(x^2+y^2)^2}\,dx\,dy=\int_{0}^{1}\int_{1-x}^{\sqrt{1-x^2}} \frac{1}{(x^2+y^2)^2}\,dy\,dx .$$