<s>[INST] 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? 
 [/INST] $\newcommand{\I}{\mathfrak{I}}$$\newcommand{\dx}{\mathrm dx\,}$Let’s take the time to generalize this. Denote the generalized integral as$$\I=\int\limits_0^{\infty}\dx\frac 1{x^4+ax^2+b^2}$$Now factor out an $x^2$ from the denominator and complete the square to get
$$\I=\int\limits_0^{\infty}\frac {\dx}{x^2}\frac 1{\left(x-\frac bx\right)^2+a+2b}$$Make the transformation $x\mapsto\tfrac bx$ so that$$\I=\frac 1b\int\limits_0^{\infty}\dx\frac 1{\left(x-\frac bx\right)^2+a+2b}$$Therefore, it’s easy to see that
$$b\I=\int\limits_0^{\infty}\frac {\dx}{x^2}\frac b{\left(x-\frac bx\right)^2+a+2b}=\int\limits_0^{\infty}\dx\frac 1{\left(x-\frac bx\right)^2+a+2b}$$
Add the two integrals together to get
$$\begin{align*}\I & =\frac 1{2b}\int\limits_0^{\infty}\dx\left(1+\frac b{x^2}\right)\frac 1{\left(x-\frac bx\right)^2+a+2b}\\ & =\frac 1{2b}\int\limits_{-\infty}^{\infty}\frac {\dx}{x^2+a+2b}\\ & \color{blue}{=\frac {\pi}{2b\sqrt{a+2b}}}\end{align*}$$
Now set $a=7$ and $b=1$ to get the answer.
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