Who's Up To the Challenge? 84

$A=\int_0^{\infty}\frac{1}{x^4+x^2+1}dx$

Now, making a change of variables $\displaystyle x=\frac{1}{t}$ implies $\displaystyle dx=-\frac{1}{t^2}dt$

Thus,

$A=-\int_{\infty}^0\frac{t^2}{t^4+t^2+1}dt$

$=\int_0^{\infty}\frac{x^2}{x^4+x^2+1}dx$

$=\frac{1}{2}\int_{-\infty}^{\infty}\frac{1}{\left(x-\frac{1}{x}\right)^2+3}dx$

Now, using the Cauchy Schlomilch transformation,

$A=\frac{1}{2}\int_{-\infty}^{\infty}\frac{1}{x^2+3}dx$

$=\frac{\pi}{2\sqrt{3}}$

Thus, $\displaystyle a=1\ ,\ b=2\ ,\ c=3$ making the answer $\displaystyle \boxed{6}$

We have $\begin{aligned} A & = \; \int_0^\infty \frac{1}{x^4+x^2+1}\,dx \; =\; \tfrac12\int_{\mathbb{R}} \frac{1}{x^4 + x^2 + 1}\,dx \; = \; \tfrac12\int_{\mathbb{R}}\frac{1}{(x^2-\omega)(x^2-\omega^2)}\,dx \\ & = \; \pi i \Big(\mathrm{Res}_{z=\omega} + \mathrm{Res}_{z=-\omega^2}\Big) \frac{1}{(z^2-\omega)(z^2-\omega^2)} \\ & = \; \pi i\left(\frac{1}{2\omega(\omega^2-\omega)} + \frac{1}{-2\omega^2(\omega-\omega^2)}\right) \; = \; \frac{\pi}{2\sqrt{3}} \end{aligned}$ making the answer $1 + 2 + 3 = \boxed{6}$ .