Double integral over an annulus

Polar. $I=\int_0^{2\pi}\int_2^4\frac{r\,dr\,d\theta}{r^2(1+\cos\theta\sin\theta)}=\int_2^4\frac{dr}{r}\cdot\int_0^{2\pi}\frac{d\theta}{1+\cos\theta\sin\theta}$.

Radial. $\int_2^4 dr/r=\ln2$.

Angular integral. Using $\cos\theta\sin\theta=\frac12\sin 2\theta$ and $t=\theta-\pi/2$, substitution $u=\tan t$:
$$J=\int_0^{2\pi}\frac{d\theta}{1+\frac12\sin2\theta}=2\int_{-\pi/2}^{\pi/2}\frac{dt}{1-\frac12\sin2t}=2\int_{-\infty}^{+\infty}\frac{du}{(u-\frac12)^2+\frac34}=\frac{4\pi}{\sqrt3}.$$

Total. $I=\frac{4\pi\ln2}{\sqrt3}$.

Conclusion. $I=\frac{4\pi}{\sqrt3}\ln2$.