Integral of $1/(x^2 + y^2)^2$ under the parabola $y = x^2 - x - 1$ I'm trying to show that
\begin{equation}
 \int \limits_{y < x^2 - x - 1} \!\!\!\! \frac{1}{(x^2 + y^2)^2}\, dxdy = \pi,
 \tag{1}
\end{equation}
and more generally that
\begin{equation}
 \int\limits_{y < ax^2 + bx + c} \!\!\!\! \frac{1}{(x^2 + y^2)^2}\, dxdy
 = \left(\frac{b^2 + 1 - 2ac}{4c^2}\right) \pi,
 \tag{2}
\end{equation}
for $a > 0$ and $c < 0$. I believe these equations are true based on numerical computations.
It is pretty straightforward to integrate with respect to $y$ using the substitution $y = x\tan(v)$, but that turns the integral in equation (1) into
\begin{equation}
    \frac{1}{2} \int \limits_{-\infty}^{\infty} \frac{1}{x^3} \left(\arctan(x - 1 - 1/x) + \mathrm{sgn}(x)\frac{\pi}{2} + \frac{x - 1 - 1/x}{1 + (x - 1 - 1/x)^2}\right)dx,
\end{equation}
which looks very difficult. I hope there is a more clever approach to the original integral.
 A: We can factor the parabola into the form $y=a(x-A)(x-B)$.  If $A=B$ or $A$ and $B$ are not real numbers, out integral will diverge.  If $AB<0$, then polar substitution converts the integral into the form of $\iint\frac{dr}{r^3}d\theta$.  The limits of integration are from $r=0$ to $\frac{\sin(\theta)-b\cos(\theta)+\sqrt{\sin^2(\theta)-2b\sin(\theta)\cos(\theta)+b^2\cos^2(\theta)-4ac\cos^2(\theta)}}{2a\cos^2(\theta)}$ and either from $\theta=0$ to $\theta=\pi$ or from $\theta=\pi$ to $\theta=2\pi$, depending on the sign of $a$.  If $AB>0$, then the same substitution can be done, with the limits of integration from $r=\frac{\sin(\theta)-b\cos(\theta)-\sqrt{\sin^2(\theta)-2b\sin(\theta)\cos(\theta)+b^2\cos^2(\theta)-4ac\cos^2(\theta)}}{2a\cos^2(\theta)}$ to $r=\frac{\sin(\theta)-b\cos(\theta)+\sqrt{\sin^2(\theta)-2b\sin(\theta)\cos(\theta)+b^2\cos^2(\theta)-4ac\cos^2(\theta)}}{2a\cos^2(\theta)}$.  The limits with respect to $\theta$ are based on the location of the vertex of the parabola, and either begin or end at $\theta=0$ or $\theta=\pi$.
Depending on the values of $a,b,c$, this integral might be easier to solve
