Fun with 2018 #5

$\begin{aligned} 2018 & = \int_0^\infty \frac {4A\ dx}{2018\left(e^\frac x{2018}+e^{-\frac x{2018}}\right)^2} \\ & = \frac A{2018} \int_0^\infty \text{sech}^2 \left(\frac x{2018}\right) dx & \small \color{#3D99F6} \text{Let }u = \frac x{2018} \implies du = \frac {dx}{2018} \\ & = A \int_0^\infty \text{sech}^2 u \ du \\ & = A \tanh u \ \bigg|_0^\infty \\ & = A \end{aligned}$

Therefore, $A=\boxed{2018}$ .