Evaluating $I=\int_{-\infty}^\infty \frac{u^2}{5u^2\left(u^2+1\right)+2}\,\mathrm{d}u$? This integral popped up when I was trying to solve this integral : $$\int_{-\pi/2}^{\pi/2}\frac{\sin^2x}{4(\cos^4x+2\sin^4x)+2\sin^2 2x}\,dx $$
I simplified it a little bit, substituted $\tan(x)=u$ and came up with, 
$$I= \int_{-\infty}^\infty \frac{u^2}{5u^2\left(u^2+1\right)+2}\,\mathrm{d}u$$
any suggestions for how I can take it from here ?

If we assume, as in mickep's comment, that it's $\dfrac{u^2}{8u^4 + 8u^2+4},$ then we work first on factoring the denominator:
$$
8u^4 + (\cdots) + 4 = \Big( \sqrt8\, u^2 + 2\Big)^2 = 8u^4 + 8\sqrt 2\, u^2 + 2
$$
So
\begin{align}
8u^4 + 8u^2 + 4 & = \Big(8u^2 + 8\sqrt 2 u^2 + 2\Big) + (8u^2 - 8\sqrt 2\, u^2) \\[10pt]
& = (\sqrt8\,u^2 + 2)^2 + 8(1-\sqrt2)u^2 \\[10pt]
& = \Big( \sqrt8\,u^2 + 2\Big)^2 - \Big( \sqrt{8(\sqrt 2 - 1)} \, u \Big)^2 \\[10pt]
& = \Big( \sqrt8\, u^2 + 1 - \sqrt{8(\sqrt 2 - 1)}\,u\, \Big) \Big( \sqrt8\, u^2 + 1 + \sqrt{8(\sqrt 2 - 1)}\,u \, \Big).
\end{align}
You then have a product of two irreducible quadratic factors in the denominator. Do partial fractions accordingly.
You might also consider the tangent half-angle substitution:
$$
w = \tan \frac x 2
$$
which leads to
\begin{align}
\sin x & = \frac{2w}{1+w^2}, \\[10pt]
\cos x & = \frac{1-w^2}{1+w^2}, \\[10pt]
dx & = \frac{2\,dw}{1+w^2}.
\end{align}
