Answer
We show that
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} {\left( {{x^2} + {y^2} + {z^2} + 1} \right)^{ - 2}}{\rm{d}}V = {\pi ^2}$,
where ${\cal W}$ denotes the region over all of ${\mathbb{R}^3}$.
Work Step by Step
We have $f\left( {x,y,z} \right) = {\left( {{x^2} + {y^2} + {z^2} + 1} \right)^{ - 2}}$.
In spherical coordinates:
$f\left( {\rho \sin \varphi \cos \theta ,\rho \sin \varphi \sin \theta ,\rho \cos \varphi } \right) = {\left( {{\rho ^2} + 1} \right)^{ - 2}}$
Let ${\cal W}$ denote the region over all of ${\mathbb{R}^3}$.
We evaluate the triple integral of ${\left( {{x^2} + {y^2} + {z^2} + 1} \right)^{ - 2}}$ over all of ${\mathbb{R}^3}$ using spherical coordinates:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} {\left( {{x^2} + {y^2} + {z^2} + 1} \right)^{ - 2}}{\rm{d}}V = \mathop {\lim }\limits_{R \to \infty } \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\varphi = 0}^\pi \mathop \smallint \limits_{\rho = 0}^R {\left( {{\rho ^2} + 1} \right)^{ - 2}}{\rho ^2}\sin \varphi {\rm{d}}\rho {\rm{d}}\varphi {\rm{d}}\theta $
(1) ${\ \ \ }$ $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} {\left( {{x^2} + {y^2} + {z^2} + 1} \right)^{ - 2}}{\rm{d}}V = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\varphi = 0}^\pi \left( {\mathop {\lim }\limits_{R \to \infty } \mathop \smallint \limits_{\rho = 0}^R \frac{{{\rho ^2}}}{{{{\left( {1 + {\rho ^2}} \right)}^2}}}{\rm{d}}\rho } \right)\sin \varphi {\rm{d}}\varphi {\rm{d}}\theta $
Consider the inner integral: $\mathop \smallint \limits_{\rho = 0}^R \frac{{{\rho ^2}}}{{{{\left( {1 + {\rho ^2}} \right)}^2}}}{\rm{d}}\rho $
Write $\rho = \tan u$. So, $d\rho = {\sec ^2}udu$. Since $1 + {\tan ^2}u = {\sec ^2}u$, so
$\frac{{{\rho ^2}}}{{{{\left( {1 + {\rho ^2}} \right)}^2}}}d\rho = \frac{{{{\tan }^2}u}}{{{{\sec }^4}u}}{\sec ^2}udu = {\sin ^2}udu$
Thus,
$\mathop \smallint \limits_{\rho = 0}^R \frac{{{\rho ^2}}}{{{{\left( {1 + {\rho ^2}} \right)}^2}}}{\rm{d}}\rho = \mathop \smallint \limits_{u = 0}^{{{\tan }^{ - 1}}R} {\sin ^2}u{\rm{d}}u$
$ = \mathop \smallint \limits_{u = 0}^{{{\tan }^{ - 1}}R} {\sin ^2}u{\rm{d}}u$
Since ${\sin ^2}u = \frac{{1 - \cos 2u}}{2}$, so
$\mathop \smallint \limits_{\rho = 0}^R \frac{{{\rho ^2}}}{{{{\left( {1 + {\rho ^2}} \right)}^2}}}{\rm{d}}\rho = \frac{1}{2}\mathop \smallint \limits_{u = 0}^{{{\tan }^{ - 1}}R} \left( {1 - \cos 2u} \right){\rm{d}}u$
$ = \frac{1}{2}\left( {u - \frac{1}{2}\sin 2u} \right)|_0^{{{\tan }^{ - 1}}R}$
$ = \frac{1}{2}{\tan ^{ - 1}}R - \frac{1}{4}\sin \left( {2{{\tan }^{ - 1}}R} \right)$
Applying the limit, we get:
$\mathop {\lim }\limits_{R \to \infty } \mathop \smallint \limits_{\rho = 0}^R \frac{{{\rho ^2}}}{{{{\left( {1 + {\rho ^2}} \right)}^2}}}{\rm{d}}\rho = \mathop {\lim }\limits_{R \to \infty } \left( {\frac{1}{2}{{\tan }^{ - 1}}R - \frac{1}{4}\sin \left( {2{{\tan }^{ - 1}}R} \right)} \right)$
As $R \to \infty $, we have ${\tan ^{ - 1}}R \to \frac{\pi }{2}$, so
$\mathop {\lim }\limits_{R \to \infty } \mathop \smallint \limits_{\rho = 0}^R \frac{{{\rho ^2}}}{{{{\left( {1 + {\rho ^2}} \right)}^2}}}{\rm{d}}\rho = \frac{\pi }{4}$
Submitting this result in equation (1), we obtain
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} {\left( {{x^2} + {y^2} + {z^2} + 1} \right)^{ - 2}}{\rm{d}}V = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\varphi = 0}^\pi \left( {\mathop {\lim }\limits_{R \to \infty } \mathop \smallint \limits_{\rho = 0}^R \frac{{{\rho ^2}}}{{{{\left( {1 + {\rho ^2}} \right)}^2}}}{\rm{d}}\rho } \right)\sin \varphi {\rm{d}}\varphi {\rm{d}}\theta $
$ = \frac{\pi }{4}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{\varphi = 0}^\pi \sin \varphi {\rm{d}}\varphi {\rm{d}}\theta $
$ = - \frac{\pi }{4}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\cos \varphi |_0^\pi } \right){\rm{d}}\theta $
$ = - \frac{\pi }{4}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( { - 1 - 1} \right){\rm{d}}\theta $
$ = \frac{\pi }{2}\mathop \smallint \limits_{\theta = 0}^{2\pi } {\rm{d}}\theta = {\pi ^2}$
Thus, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} {\left( {{x^2} + {y^2} + {z^2} + 1} \right)^{ - 2}}{\rm{d}}V = {\pi ^2}$.
