Evaluate: $u =\int_0^\infty\frac{dx}{x^4 +7x^2+1}$ Evaluate: 

$u =\displaystyle\int_0^\infty\dfrac{dx}{x^4 +7x^2+1}$

Attempt: 
$$u = \int_0^\infty \dfrac{dx}{\left(x^2+ \left(\dfrac{7 - \sqrt {45}}{2}\right)\right)\left(x^2+ \left(\dfrac{7 + \sqrt {45}}{2}\right)\right)}$$
$$u = \int_0^\infty \dfrac{dx}{(x^2+a^2)(x^2+b^2)}$$
After partial fraction decomposition and simplifying I get: 
$u = \dfrac{\pi}{2(a+b)}$
But answer is $\frac \pi 6$.
Where have I gone wrong? 

A slightly different approach: you have $a^2+b^2=7$ and $a^2 b^2=1$ with $a,b\in\mathbb{R}^+$, such that
$$ \operatorname*{Res}_{x=ia}\frac{1}{(x^2+a^2)(x^2+b^2)} = \lim_{x\to ia}\frac{1}{(x+ia)(x^2+b^2)} = \frac{1}{2i}\cdot \frac{1}{a(b^2-a^2)}$$
and similarly
$$ \operatorname*{Res}_{x=ib}\frac{1}{(x^2+a^2)(x^2+b^2)} = \lim_{x\to ib}\frac{1}{(x+ib)(x^2+a^2)} = \frac{1}{2i}\cdot \frac{1}{b(a^2-b^2)}$$
so $\pi i$ times the sum of the residues at $ia$ and $ib$ equals $\frac{\pi}{2ab(a+b)}$. On the other hand the integral is blatantly positive, such that $ab(a+b)$ is the square root of $a^2 b^2(a^2+b^2+2\sqrt{a^2 b^2}) = 9$ and voilà the wanted outcome $\frac{\pi}{6}$.
