Integrate I guess

$\begin{aligned} I & = \int_{\frac {2e}{e^2-1}}^\infty \frac {\color{#3D99F6}\sin \left(\tan^{-1} x\right)}{x^4} dx \\ & = \int_{\frac {2e}{e^2-1}}^\infty \frac {\color{#3D99F6}x}{x^4\color{#3D99F6}\sqrt{x^2+1}} dx \\ & = \int_{\frac {2e}{e^2-1}}^\infty \frac 1{x^3\sqrt{x^2+1}} dx & \small \color{#3D99F6} \text{Let }x = \tan u \implies dx = \sec^2 u \ du \\ & = \int_{\color{#3D99F6}\alpha}^\frac \pi 2 \frac {\sec u}{\tan^3 u}du & \small \color{#3D99F6} \text{where } \alpha = \tan^{-1} \frac {2e}{e^2-1} \\ & = \int_\alpha^\frac \pi 2 \frac {\cos^2 u}{\sin^3 u}du \\ & = \int_\alpha^\frac \pi 2 \frac {1-\sin^2 u}{\sin^3 u}du \\ & = {\color{#3D99F6}\int_\alpha^\frac \pi 2 \csc^3 u\ du} - \int_\alpha^\frac \pi 2 \csc u\ du & \small \color{#3D99F6} \text{By reduction formula} \\ & = {\color{#3D99F6}- \frac {\cos x \csc^2 x}2 \bigg|_\alpha^\frac \pi 2 + \frac 12 \int_\alpha^\frac \pi 2 \csc u\ du} - \int_\alpha^\frac \pi 2 \csc u\ du \\ & = \frac {e^4-1}{8e^2} - {\color{#3D99F6}\frac 12 \int_\alpha^\frac \pi 2 \csc u\ du} & \small \color{#3D99F6} \text{Multiply up and down by }\csc u + \cot u \\ & = \frac {e^2-e^{-2}}8 - {\color{#3D99F6}\frac 12 \int_\alpha^\frac \pi 2 - \frac {-\csc^2 u - \csc x \cot x}{\csc u + \cot u}\ du} \\ & = \frac {\sinh 2}4 + \frac 12 \ln (\csc u + \cot u) \bigg|_\alpha^\frac \pi 2 \\ & = \frac {\sinh 2}4 - \frac 12 \\ & = \frac {\sinh 2-2}4 \end{aligned}$

$\implies a+b+c = 4+2+2 = \boxed{8}$