Integral Impropria

Basta usar coordenadas polares:

\( \begin{cases}
x = r \cos \theta \\
y =  r \sin \theta
\end{cases}
 \implies dxdy  = r dr d\theta \)


Logo
\( \begin{array}{cl}
I &=\displaystyle \int_{\mathbb R^2} \dfrac{1}{x^2 + y^2 + 1^{3/2}} dxdy  \\[2ex]
 &=\displaystyle \int_0^{2\pi} \int_0^\infty \dfrac{1}{\left( (r\cos \theta)^2 + (r \sin \theta)^2 + 1\right)^{3/2}} r dr d\theta \\[2ex]
 & = \displaystyle \int_0^{2\pi} d\theta  \int_0^\infty \dfrac{r}{(r^2+ 1)^{3/2}}  dr = 2 \pi \int_0^\infty \dfrac{r}{(r^2+ 1)^{3/2}}  dr
\end{array} \)

Para a última integral, use a substituição \(t = r^2 + 1\implies dt = 2rdr \)


\( I= \pi \displaystyle \int_1^\infty  \dfrac{ 1}{t^{3/2}} dt = -\pi \cdot \dfrac{2}{t^{1/2}}\Bigg|_{t = 1}^{t = \infty} = 2 \pi \)