Classic Calculus Question I

Performing the integral using spherical polar coordinates, and the substitution $u = r^2$ , $\begin{aligned} \iiint_{\mathbb{R}^3}\frac{1}{(x^2+y^2+z^2+1)^2}\,dx\,dy\,dz & = \; \int_0^{2\pi} \int_0^\pi \int_0^\infty \frac{1}{(r^2+1)^2}\,r^2\sin\theta\,dr\,d\theta\,d\phi \; = \; 4\pi \int_0^\infty \frac{r^2}{(r^2+1)^2}\,dr \\ & = \; 2\pi \int_0^\infty \frac{\sqrt{u}}{(u+1)^2}\,du \; = \; 2\pi B(\tfrac32,\tfrac12) \; = \; 2\pi \frac{\Gamma(\tfrac32)\Gamma(\tfrac12)}{\Gamma(2)} \; = \; 2\pi \frac{\frac12\sqrt{\pi} \times \sqrt{\pi}}{1!} \\ & = \; \pi^2 \end{aligned}$ making the answer $\boxed{2}$ .

To add to Mark Henning's solution, cylindrical coordinates can also do the trick here. Let $0 \le \theta \le 2\pi, 0 \le r < \infty, -\infty < z < \infty$ , and $x^2 + y^2 = r^2$ which gives us the triple integral:

$\int_{0}^{2\pi} \int_{-\infty}^{\infty} \int_{0}^{\infty} \frac{1}{(r^2 + z^2 + 1)^2} r drdzd\theta = \boxed{\pi^2}.$