<s>[INST] Evaluate $\int_0^{\infty} \frac{ dx }{(x^4+c)(x^2+1) } $ Consider $$\int_0^{\infty} \dfrac{ dx }{(x^4+c)(x^2+1) } $$
where $c \in \mathbb{R}$. Now, it is possible to evaluate this using complex analysis, but I was wondering if there is way to compute this integral without complex analysis.
I was thinking on partial fractions, but there is no easy form to decompose it too.
Here is a way I was thinking: Write $1 = \dfrac{c+x^4 - x^4 }{c}$ to obtain 
$$ \frac{1}{c} \int_0^{\infty} \frac{dx}{x^2+1}  - \dfrac{1}{c} \int_0^{\infty} \dfrac{x^4}{(x^4+c)(x^2+1)} = \dfrac{ \pi }{2c} - \dfrac{1}{c} \int_0^{\infty} \dfrac{x^4}{(x^4+c)(x^2+1)}$$
I dont know if this is "progress", what do you think?
 [/INST] $$\dfrac{ 1 }{(x^4+c)(x^2+1) }=\frac{1}{\left(x^2+1\right) \left(x^2-i \sqrt{c}\right) \left(x^2+i \sqrt{c}\right)}$$ Using partial fractions, it is
$$-\frac{1}{2 \left(\sqrt{c}-i\right) \sqrt{c} \left(x^2-i \sqrt{c}\right)}-\frac{1}{2
   \left(\sqrt{c}+i\right) \sqrt{c} \left(x^2+i \sqrt{c}\right)}-\frac{i}{2
   \left(\sqrt{c}-i\right) \left(\sqrt{c}+i\right) (x-i)}+\frac{i}{2
   \left(\sqrt{c}-i\right) \left(\sqrt{c}+i\right) (x+i)}$$ which is not so bad.
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