If \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{96 x^2 \cos ^2 x}{\left(1+e^x\right)} \mathrm{d} x=\pi\left(\alpha \pi^2+\beta\right), \alpha, \beta \in \mathbb{Z}, then (\alpha+\beta)^2 equals

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Answer: B. 100 \int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{96 x^2 \cos ^2 x}{\left(1+e^xight)} d x \quad (Apply King Property) \begin{aligned} & \int_0^{\frac{\pi}{2}} 96 x^2 \cos ^2 x=48 \int_0^{\frac{\pi}{2}} x^2(1+\cos 2 x) d x \ & 48\left[\left(\frac{x^3}{3}ight)_0^{\pi / 2}+\int_0^{\frac{\pi}{2}} x_1^2 \cos _{I I} 2 x d xight] \ & \Rightarrow 	ext { On solving } \pi\left(2 \pi^2-12ight) \ & \Rightarrow \alpha=2, \beta=-12 \ & \Rightarrow(\alpha+\beta)^2=100 \end{aligned}

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

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