If ${\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}} \frac{96x^2{\cos }^{2}x}{\left(1+e^x\right)}dx=\pi \left(\alpha {\pi }^2+\beta \right),\alpha ,\beta \in Z,$ then $(\alpha +\beta {)}^2$ equals

$\begin{array}{l}{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}} \frac{96x^2{\cos }^{2}x}{\left(1+e^x\right)}dx\text{ (Apply King's Property) }\end{array}$

$\begin{array}{l}{\int }_0^{\frac{\pi }{2}} 96x^2{\cos }^{2}x=48{\int }_0^{\frac{\pi }{2}} x^2(1+\cos 2x)dx\end{array}$

$48\left[{\left(\frac{x^3}{3}\right)}_0^{\pi /2}+{\int }_0^{\frac{\pi }{2}} \underset{I}{x^2} \underset{II}{\cos 2x}dx\right]$

$48\left[\left(\frac{{\pi }^3}{24}-0\right)+\left[x^2{\int }_0^{\frac{\pi }{2}}\cos 2x dx-{\int }_0^{\frac{\pi }{2}} \frac{d}{dx}\left(x^2\right) {\int }_0^{\frac{\pi }{2}}\cos 2xdx\right]dx\right]$

$\left[2{\pi }^3+48\left[\left(0-0\right)-{\int }_0^{\frac{\pi }{2}} \cancel{2}x \left(\frac{\sin 2x}{\cancel{2}}\right)\right]dx\right]$

$\left[2{\pi }^3+48\left[-{\int }_0^{\frac{\pi }{2}} x\sin 2x\right]dx\right]$

$=2{\pi }^3-48\left[x{\int }_0^{\frac{\pi }{2}} \sin 2x-{\int }_0^{\frac{\pi }{2}}\frac{d}{dx}\left( x\right){\int }_0^{\frac{\pi }{2}} \sin 2xdx\right]dx$

$=2{\pi }^3-48\left[{\left(x\frac{-\cos 2x}{2}\right)}_0^{\frac{\pi }{2}}-{\int }_0^{\frac{\pi }{2}}\frac{-\cos 2x}{2}\right]dx$

$=2{\pi }^3-48\left[\frac{\pi }{4}-0+{\left(\frac{\sin 2x}{4}\right)}_0^{\frac{\pi }{2}}\right]dx$

$\begin{array}{l}\Rightarrow \text{ On solving }\pi \left(2{\pi }^2-12\right)\\ \Rightarrow \alpha =2,\beta =-12\\ \Rightarrow (\alpha +\beta {)}^2=100\end{array}$