$$\int \int _R {\,dx \,dy \over \sqrt{x^2 +y^2} },$$ where $R : \{|x| \le 1; |y| \le 1 \}$.
My solutions: $$\int \int _R {\,dx \,dy \over \sqrt{x^2 +y^2} }=4\int_{\theta=0}^{\pi \over 4} \int_{r=0}^{\sqrt 2}{r \,dx \,dy \over r}=\sqrt2 \pi$$
Can we do this way?
You were on the right track with symmetry. We can say that
$$\int_{-1}^1\int_{-1}^1 \frac{dydx}{\sqrt{x^2+y^2}} = 8 \int_0^1 \int_0^x \frac{dydx}{\sqrt{x^2+y^2}}$$
Then, to integrate the triangle, we have to find the boundary $x=1$ in polar coordinates
$$1 = x = r\cos\theta \implies r = \frac{1}{\cos\theta} = \sec\theta$$
which means the correct integral is
$$8\int_0^{\frac{\pi}{4}} \int_0^{\sec\theta} drd\theta$$
Can you take it from here?