compute $\int_Q\frac{1}{|x|} \, dx$ on $Q=[0,1]^2$

$$\begin{eqnarray*}\iint_{(0,1)^2}\frac{da\,db}{\sqrt{a^2+b^2}}&\stackrel{\text{symmetry}}{=}&2\int_{0}^{1}\int_{0}^{a}\frac{1}{\sqrt{a^2+b^2}}\,db\,da\\&\stackrel{b\mapsto a c}{=}&2\int_{0}^{1}\int_{0}^{1}\frac{1}{\sqrt{1+c^2}}\,dc\,da\\&=&2\,\text{arcsinh}(1)=\color{red}{2\log(1+\sqrt{2}).}\end{eqnarray*}$$

$$0 \leq \tan\theta \leq \infty \implies \theta \in [0,\pi/2]$$

$$0 \leq r\cos\theta \leq 1, 0 \leq r\sin\theta \leq 1 \implies 0 \leq r \leq \min(1/\cos\theta,1/\sin\theta) $$

$$\begin{align}I &= \int_0^{\pi/2}\int_0^{\min(1/\cos\theta,1/\sin\theta)} \,\partial r \,\partial \theta\\ &= \int_0^{\pi/4}\int_0^{1/\cos\theta} \, \partial r \, \partial \theta + \int_{\pi/4}^{\pi/2}\int_0^{1/\sin\theta} \, \partial r \, \partial \theta \\ &= \int_0^{\pi/4}\frac{1}{\cos\theta} \, \partial \theta + \int_{\pi/4}^{\pi/2} \frac{1}{\sin\theta} \, \partial \theta \\ &= \log(\tan\theta+\sec\theta)|_0^{\pi/4} -\log(\cot\theta+\csc\theta)|_{\pi/4}^{\pi/2}\\ &= \log(1+\sqrt 2) - 0-0 + \log(1+\sqrt{2}) \\ &= 2\log(1+\sqrt 2)\end{align}$$

Note that I used the fact that $\sin\theta\leq\cos\theta, \forall \theta \in [0,\pi/4]$ and $\sin\theta>\cos\theta, \forall \theta \in (\pi/4,\pi/2]$.