Change to an equivalent polar integral and evaluate
$$\int_0^1\Biggr(\int_{-\sqrt{1-x^2}}^\sqrt{1-x^2} \frac{6}{(1+x^2+y^2)^2} dy \Biggr)dx$$
I am working on this double integral and have run into a little snag. Obviously, our best bet here is to convert to polar coordinates. Our restrictions here give us a graph that looks like a full circle with a radius of $1$; the $x$ restriction slices it in half through the $y$-axis. So in polar form, our new bounds would be from $0$ to $1$ for our $r$-values and $\pi/2$ to $-\pi/2$ for our $\theta$ bound. The system solved its self quite well using this method. But I have noticed that others on various websites are restricting their $\theta$ values from $0$ to $2\pi$.
So my question here is what should the bounds be for the $\theta$ variable?
As Alex mentioned, what you have worked out is correct, $\theta$ should be in the interval $[-\pi/2, \pi/2]$.
Let $D=\{(x,y)\in\Bbb R^2|0 \leq x \leq 1, -\sqrt{1-x^2} \leq y \leq \sqrt{1-x^2}\}$
By Fubini's Theorem we have that:
$$\int_0^1\Biggr(\int_{-\sqrt{1-x^2}}^\sqrt{1-x^2} \frac{6}{(1+x^2+y^2)^2} dy \Biggr)dx=\iint_D \frac{6}{(1+x^2+y^2)^2} dA$$
Now by the Change of Variables Theorem we change to polar coordinates
$$\iint_D \frac{6}{(1+x^2+y^2)^2} dA=\iint_{[\pi/2,\pi/2]\times[0,1]} \frac{6}{(1+r^2)^2} rdrd\theta$$
Now we apply Fubini's Theorem again and then simply evaluate the integral
$$\int_{\theta=-\pi/2}^{\pi/2}\Biggr(\int_{r=0}^1 \frac{6}{(1+r^2)^2} rdr\Biggr)d\theta=\int_{\theta=-\pi/2}^{\pi/2} \frac{3}{2} d\theta=\frac{3\pi}{2}$$
Note that because of the symmetry of the function $f(x,y) = \frac{6}{(1+x^2+y^2)^2}$ we could have used the limits $0$ and $2\pi$ for $\theta$, however we should remember to multiply the answer by $\frac{1}{2}$ because otherwise we would have calculated double the volume required.