How to solve this triple integral? I am wondering how I could solve the integral 
$$\iiint \frac{1-e^{-(x^2+y^2+z^2)}}{[x^2+y^2+z^2]^{2}}$$
over $\mathbb{R}^{3}$
I thought maybe I could break it up into three single integrals and multiply or something. I think it is not supposed to be difficult to solve. How should it be approached?
Thank you
 A: Hint: Use spherical coordinates, and note that each copy of $\mathbb{R} = \left(-\infty,\infty\right)\Rightarrow I = \displaystyle \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \dfrac{1-e^{-\left(x^2+y^2+z^2\right)}}{(x^2+y^2+z^2)^2}dxdydz= 8\displaystyle \lim_{r \to \infty} \int_{0}^r \int_{0}^r \int_{0}^r \dfrac{1-e^{-\left(x^2+y^2+z^2\right)}}{(x^2+y^2+z^2)^2}dxdydz$. Can you take it from here ?
A: Hint. By the use of a symetry and by use of spherical coordinates 
$$
\begin{align}
x&=r\sin\theta\cos\phi\\
y&=r\sin\theta\sin\phi\\
z&=r\cos\theta
\end{align}
$$ one gets a jacobian equal to $r^2\sin \theta \:d\phi \:d\theta\: dr$ giving
$$
\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \dfrac{1-e^{-\left(x^2+y^2+z^2\right)}}{(x^2+y^2+z^2)^2}dxdydz=8\int_0^{\pi/2}\int_0^{\pi/2}\int_0^\infty\dfrac{1-e^{-r^2}}{r^2}dr \sin \theta \:d\phi \:d\theta
$$ Integrating by parts and using the gaussian integral gives
$$
\int_0^\infty\dfrac{1-e^{-r^2}}{r^2}dr =\sqrt{\pi}
$$ leading to a conclusion.
