I have $$D=\{(x,y,z): x^2<z^2<x^2+y^2\wedge x>0\wedge y>0\wedge z<x^2+y^2<1\}$$ and I want to evaluate the integral $$\iiint_D\frac{y}{(x^2+y^2+z^2)\sqrt{x^2+y^2} }\, dx\, dy\, dz\,.$$
By using the cylindrical coordinates I obtain $$\int_{0}^{\pi/2} \int_{0}^{1} \int_{r\cos \theta}^{r} \frac{r \sin \theta}{z^2+r^2}\, dz\, dr\, d\theta$$ Is it correct?
If I calculate it I found $\frac{\pi}{4}-\int_{0}^{1} \frac{\arctan t}{\sqrt{1-t^2}}dt$ and this integral exists finitely.
Yes, the integral in cylindrical coordinates is written correctly. Your computation is almost right, but something happened near the end. $$\begin{align} \int_{0}^{\pi/2} \int_{0}^{1} \int_{r\cos \theta}^{r} \frac{r \sin \theta}{z^2+r^2} dz\, dr\, d\theta&=\int_{0}^{\pi/2} \int_{0}^{1} \sin \theta\left[\arctan(z/r)\right]_{r\cos \theta}^{r} dr\, d\theta\\ &=\int_{0}^{\pi/2} \int_{0}^{1} \sin \theta\left(\frac{\pi}{4}-\arctan(\cos(\theta))\right) dr\, d\theta\\ &=\frac{\pi}{4}+\int_{0}^{\pi/2} \arctan(\cos(\theta))(-\sin(\theta))d\theta\\&= \frac{\pi}{4}-\int_{0}^{1} \arctan(t)dt\\ &=\frac{\pi}{4}-\left[t\arctan(t)-\frac{1}{2}\ln(1+t^2)\right]_{0}^{1}=\frac{\ln(2)}{2}. \end{align}$$