$\int_0^1 \frac{1}{\left(5+2 x-2 x^2\right)\left(1+e^{(2-4 x)}\right)} d x=\frac{1}{\alpha} \log _e\left(\frac{\alpha+1}{\beta}\right)$. $\alpha, \beta,0$, then $\alpha^4-\beta^4$ is equal to: [JEE Main 2023, 15 Apr (Shift 1)]

Question

$\int_0^1 \frac{1}{\left(5+2 x-2 x^2\right)\left(1+e^{(2-4 x)}\right)} d x=\frac{1}{\alpha} \log _e\left(\frac{\alpha+1}{\beta}\right)$. $\alpha, \beta,>0$, 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+2 x-2 x^2\right)\left(1+e^{(2-4 x)}\right)} d x=\frac{1}{\alpha} \log _e\left(\frac{\alpha+1…

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