Infinite Integral

Consider $V$ as follows:

$\begin{aligned} V & = \int_0^\infty \frac {x^2}{x^4+7x^2+1} dx & \small \color{#3D99F6} \text{Using }\int_0^\infty f(x)\ dx = \int_0^\infty \frac {f\left(\frac 1x\right)}{x^2}\ dx \\ & = \int_0^\infty \frac {\frac 1{x^2}}{x^2\left(\frac 1{x^4}+\frac 7{x^2}+1\right)} dx \\ & = \int_0^\infty \frac 1{1+7x^2+x^4} dx = U \end{aligned}$

Now consider $U$ :

$\begin{aligned} U & = \int_0^\infty \frac 1{x^4+7x^2+1} dx \\ & = \int_0^\infty \frac 1{\left(x^2+\frac {7-3\sqrt 5}2\right)\left(x^2+\frac {7+3\sqrt 5}2\right)} dx \\ & = \frac 1{3\sqrt 5} \int_0^\infty \left(\frac 1{x^2+\frac {7-3\sqrt 5}2} - \frac 1{x^2+\frac {7+3\sqrt 5}2} \right) dx \\ & = \frac 1{3\sqrt 5} \left[\frac {\tan^{-1}\frac x{\sqrt{\frac {7-3\sqrt 5}2}}}{\sqrt{\frac {7-3\sqrt 5}2}} -\frac {\tan^{-1}\frac x{\sqrt{\frac {7+3\sqrt 5}2}}}{\sqrt{\frac {7+3\sqrt 5}2}} \right]_0^\infty \\ & = \frac \pi {6\sqrt 5} \left(\frac 1{\sqrt{\frac {7-3\sqrt 5}2}} -\frac 1{\sqrt{\frac {7+3\sqrt 5}2}} \right) \\ & = \frac \pi {6\sqrt 5} \color{#3D99F6} \left(\sqrt{\frac {7+3\sqrt 5}2} -\sqrt{\frac {7-3\sqrt 5}2} \right) & \small \color{#3D99F6} \text{Note that } \left(\sqrt{\frac {7+3\sqrt 5}2} -\sqrt{\frac {7-3\sqrt 5}2} \right)^2 = 5 \\ & = \frac \pi{6\sqrt 5}\color{#3D99F6}(\sqrt 5) \\ & = \frac \pi 6 \end{aligned}$

Therefore, $UV = U^2 = \dfrac {\pi^2}{36} \implies ab+a+b = 2(72) + 2 + 36 = \boxed{110}$ .