integral computation $\int_{-\infty}^{\infty} \frac{1}{(1+x+x^2)^2} dx $ Compute the following integral: 
$\int_{-\infty}^{\infty} \frac{1}{(1+x+x^2)^2} dx$. Can some one give me some hints on how to do this?
I tried writing $(1+x+x^2)=f(x)$ and then multiplying and dividing by $\frac{d f(x)}{dx}=1+2x$ so we get $\int_{-\infty}^{\infty} \frac{1}{1+2x} \frac{1+2x}{(f(x))^2} dx $ = $\int_{-\infty}^{\infty} \frac{1}{1+2x} \frac{df(x)}{(f(x))^2} $. Here I tried to use integration by parts but I can't get anywhere.

Let
$$\mathcal{I}(\lambda) = \int_{-\infty}^{\infty} \frac{1}{x^2+x+\lambda}\;{dx} $$
where
$\lambda > 0$.  We have
$$x^2+x+\lambda = (x+\frac{1}{2})^2+(\lambda-\frac{1}{4})\ ,$$
thus by setting $t = {x+\frac{1}{2}}$ we have
$$\mathcal{I}(\lambda) = \frac{2}{\sqrt{4\lambda-1}}\int_{-\infty}^{\infty}\frac{1}{1+t^2}\;{dt} = \frac{2\pi}{\sqrt{4\lambda-1}}\ .$$
Differentiating we get
$$\mathcal{I}'(\lambda) = -\frac{4\pi}{\sqrt{(4\lambda-1)^3}}$$
and
$$-\mathcal{I}'(1) =  \frac{4\pi}{3\sqrt{3}}\ .$$
See here if you haven't seen this method before. 
