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

If $I(m, n)=\int_0^1 x^{m-1}(1-x)^{n-1} d x, m, n>0$, then $I(9,14)+I(10,13)$ is

If $\mathrm{I}=\int_0^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}} x}{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x} \mathrm{~d} x$, then $\int_0^{2I} \frac{x \sin x \cos x}{\sin ^4 x+\cos ^4 x} \mathrm{~d} x$ equals :

The value of $\int_{e^2}^{e^4} \frac{1}{x}\left(\frac{e^{\left(\left(\log _e x\right)^2+1\right)^{-1}}}{e^{\left(\left(\log _e x\right)^2+1\right)^{-1}}+e^{\left(\left(6-\log _e x\right)^2+1\right)^{-1}}}\right) d x$ is