$$\int_{-\infty}^{\infty} \frac{\text{d}x}{ax^2+bx+c} = \frac{\pi}{\sqrt{\det(A)}}$$
where $A = \begin{bmatrix}a&\frac{b}{2}\\\frac{b}{2}&c\end{bmatrix}$
The connection to matrix quadratic form is given via the equivalence:
$$ax^2 + bx + c \equiv \begin{bmatrix}x&1\end{bmatrix}\begin{bmatrix}a&\frac{b}{2}\\\frac{b}{2}&c\end{bmatrix} \begin{bmatrix}x\\1\end{bmatrix} $$
What standard results from Linear Algebra make this integral a linear transformation of the standard integral?
$$\int_{-\infty}^{\infty} \frac{\text{d}x}{x^2 + 1} = \pi$$
Note: I am relatively new here, so if I didn't interpret your actual question correctly, sorry in advance.
The integral in question is $$I=\int_{-\infty}^{\infty} \frac{\text{d}x}{ax^2+bx+c}.$$ Completing the square for the denominator yields $$I=\frac{1}{a}\int_{-\infty}^{\infty} \frac{\text{d}x}{(x+\frac{b}{2a})^2+\frac{4ac-b^2}{4a^2}}.$$ Factoring out $\frac{4ac-b^2}{4a^2}$ from the denominator gives $$I=\frac{1}{a} \frac{1}{\frac{4ac-b^2}{4a^2}}\int_{-\infty}^{\infty} \frac{\text{d}x}{\frac{4a^2}{4ac-b^2}(x+\frac{b}{2a})^2+1}=\frac{4a}{4ac-b^2} \int_{-\infty}^{\infty} \frac{\text{d}x}{\frac{4a^2}{4ac-b^2}(x+\frac{b}{2a})^2+1}.$$ Let $\frac{4a^2}{4ac-b^2}(x+\frac{b}{2a})^2=\tan^2(u)\rightarrow x=\frac{\sqrt{4ac-b^2}}{2a}\tan(u)-\frac{b}{2a}$ such that $\text{d}x=\frac{\sqrt{4ac-b^2}}{2a}(\tan^2(u)+1)\text{d}u$, where $u\in(-\frac{\pi}{2},\frac{\pi}{2})$: $$I=\frac{4a}{4ac-b^2}\frac{\sqrt{4ac-b^2}}{2a}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \text{d}u=\frac{2\pi}{\sqrt{4ac-b^2}}.$$ The determinant of A is given by $$\det(A)=ac-(\frac{b}{2})^2=\frac{4ac-b^2}{4}.$$ Thus, it follows that $$\frac{\pi}{\sqrt{\det(A)}}=\frac{2\pi}{\sqrt{4ac-b^2}}.$$ This is equal to the previously derived result.