Showing $\int_{-\infty}^{\infty}\frac {\mathrm d x} {1+x+x^2}=3 \int_0^\infty \frac {\mathrm d x} {1+x+x^2}$ Is there an easy way to see why
$$\int_{-\infty}^{\infty}\frac {\mathrm d x} {1+x+x^2}=3 \int_0^\infty \frac {\mathrm d x} {1+x+x^2}$$
without having to evaluate the integrals explicitly? I was trying substitutions but nothing worked.

Let $\displaystyle x=\frac{1}{v}$.
$$\int_{-1}^0\frac{dx}{1+x+x^2}=\int_{-1}^{-\infty}\frac{\frac{-1}{v^2}dv}{1+\frac{1}{v}+\frac{1}{v^2}}=\int_{-\infty}^{-1}\frac{dv}{1+v+v^2}$$
Let $\displaystyle x=-1-u$.
$$\int_{-\infty}^{-1}\frac{dx}{1+x+x^2}=\int_\infty^0\frac{-du}{1+(-1-u)+(-1-u)^2}=\int_0^{\infty}\frac{du}{1+u+u^2}$$
Therefore,
\begin{align*}
\int_{-\infty}^{\infty}\frac{dx}{1+x+x^2}&=\int_{-\infty}^{-1}\frac{dx}{1+x+x^2}+\int_{-1}^{0}\frac{dx}{1+x+x^2}+\int_{0}^{\infty}\frac{dx}{1+x+x^2}\\
&=3\int_{0}^{\infty}\frac{dx}{1+x+x^2}
\end{align*}
