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?

Assume $c>0$ for convergence.
$$I = \int_0^{\infty} \dfrac{ dx }{(x^4+c)(x^2+1) } 
= \frac1{c+1}\int_0^{\infty}\left( \frac{ 1}{x^2+1}- \frac{ x^2-1}{x^4+c }\right) dx \\
= \frac{\pi}{2(c+1)} - \frac{1}{c+1}\int_0^{\infty}\frac{ x^2-1}{x^4+c } dx$$
where
\begin{align}
\int_0^{\infty}\frac{ x^2-1}{x^4+c } dx 
& =\frac{\sqrt c+1}{2\sqrt c} \int_0^{\infty}\frac{ 1-\frac{\sqrt c}{x^2}}{x^2+\frac c{x^2} } dx
+ \frac{\sqrt c-1}{2\sqrt c} \int_0^{\infty}\frac{ 1+\frac{\sqrt c}{x^2}}{x^2+\frac c{x^2} } dx \\
& =\frac{\sqrt c+1}{2\sqrt c} \int_0^{\infty}\frac{d(x+\frac{\sqrt c}{x})}{(x+\frac {\sqrt c}{x} )^2-2\sqrt c}
+ \frac{\sqrt c-1}{2\sqrt c} \int_0^{\infty}\frac{d(x-\frac{\sqrt c}{x})}{(x-\frac {\sqrt c}{x} )^2+2\sqrt c} \\
& =0+  \frac{(\sqrt c-1)\pi}{(2\sqrt c)^{3/2}} 
\end{align}
Thus,
$$I = \frac{\pi}{2(c+1)} - \frac{1}{c+1}\frac{(\sqrt c-1)\pi}{(2\sqrt c)^{3/2}} 
=\frac{\pi}{2(c+1)} \left( 1- \frac{\sqrt c-1}{\sqrt2 c^{3/4}} \right)$$
