${\int }_{0}^{1}\frac{1}{\left(5+2x-2{x}^{2}\right)\left(1+{e}^{(2-4x)}\right)}dx=\frac{1}{\alpha }{\log }_{e}\left(\frac{\alpha +1}{\beta }\right).$ $\alpha ,\beta ,0$, then ${\alpha }^{4}-{\beta }^{4}$ is equal to: [JEE Main 2023, 15 Apr (Shift 1)]

Question

${\int }_{0}^{1}\frac{1}{\left(5+2x-2{x}^{2}\right)\left(1+{e}^{(2-4x)}\right)}dx=\frac{1}{\alpha }{\log }_{e}\left(\frac{\alpha +1}{\beta }\right).$ $\alpha ,\beta ,>0$, then ${\alpha }^{4}-{\beta }^{4}$ is equal to: [JEE Main 2023, 15 Apr (Shift 1)]

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\({\int }_{0}^{1}\frac{1}{\left(5+2x-2{x}^{2}\right)\left(1+{e}^{(2-4x)}\right)}dx=\frac{1}{\alpha }{\log }_{e}\left(\fr…

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