Answer
$$2\pi(2-\sqrt{e})$$
Work Step by Step
Our aim is to integrate the integral as follows:
$$\int^{2\pi}_0 \int^{\sqrt{e}}_1 (\dfrac{\ln (r^2)}{r}) \space r \space dr \space d\theta =\int^{2\pi}_0 \int^{\sqrt{e}}_1 2\ln r \space dr \space d\theta \\=2 \times \int^{2\pi}_0[r \times \ln(r)-r]^{1/2}_1 \space d\theta \\=(2) \times \int^{2\pi}_0 \sqrt{r}[\dfrac{1}{2}-1+1] \space d\theta \\=2\pi(2-\sqrt{e})$$
