If ∫ _ 0 ^ 1 1 (5 + 2x - 2 x ^ 2 )(1 + e ^ (2 - 4x) ) dx = (1)/(α)...

Given,

${∫}_{0}^{1}\dfrac{1}{\left(5 + 2x - 2{x}^{2}\right)\left(1 + {e}^{\left(2 - 4x\right)}\right)}dx = \dfrac{1}{\alpha}\log_{e}\left(\dfrac{\alpha + 1}{\beta}\right)$

Now let,

$I = \dfrac{1}{2}{∫}_{0}^{1}\dfrac{1}{\left(\dfrac{5}{2} - \left({x}^{2} - x\right)\right)\left(1 + {e}^{- 4\left(x - \dfrac{1}{2}\right)}\right)}dx$

$⇒ I = \dfrac{1}{2}{∫}_{0}^{1}\dfrac{1}{\left(\left(\dfrac{11}{4}\right) - {\left(x - \dfrac{1}{2}\right)}^{2}\right)\left(1 + {e}^{- 4\left(x - \dfrac{1}{2}\right)}\right)}dx$

Now let $x - \dfrac{1}{2} = t ⇒ dx = dt$

So, integral becomes,

$I = \dfrac{1}{2}{∫}_{\dfrac{- 1}{2}}^{\dfrac{1}{2}}\dfrac{1}{{\left(\dfrac{\sqrt{11}}{2}\right)}^{2} - {t}^{2})\left(1 + {e}^{- 4t}\right)}dt$

$⇒ I = \dfrac{1}{2}{∫}_{\dfrac{- 1}{2}}^{0}\dfrac{1}{{\left(\dfrac{\sqrt{11}}{2}\right)}^{2} - {t}^{2})\left(1 + {e}^{- 4t}\right)}dt + \mathop{\mathop{\dfrac{1}{2}{∫}_{0}^{\dfrac{1}{2}}\dfrac{1}{{\left(\dfrac{\sqrt{11}}{2}\right)}^{2} - {t}^{2})\left(1 + {e}^{- 4t}\right)}dt}_{⏟}}_{{I}_{2}}$

Now solving ${I}_{2} = \dfrac{1}{2}{∫}_{\dfrac{- 1}{2}}^{\dfrac{1}{2}}\dfrac{1}{{\left(\dfrac{\sqrt{11}}{2}\right)}^{2} - {t}^{2})\left(1 + {e}^{- 4t}\right)}dt$,

Let $t = - z$, we get $dt = - dz$

So, ${I}_{2} = - \dfrac{1}{2}{∫}_{\dfrac{1}{2}}^{0}\dfrac{dz}{{\left(\dfrac{\sqrt{11}}{2}\right)}^{2} - {z}^{2})\left(1 + {e}^{4z}\right)}$

$⇒ {I}_{2} = \dfrac{1}{2}{∫}_{0}^{\dfrac{1}{2}}\dfrac{dt}{{\left(\dfrac{\sqrt{11}}{2}\right)}^{2} - {t}^{2})\left(1 + {e}^{4t}\right)}$

Now putting the value of ${I}_{2}\text{in}I$ we get,

$I = \dfrac{1}{2}{∫}_{0}^{\dfrac{1}{2}}\dfrac{1}{\left({\left(\dfrac{\sqrt{11}}{2}\right)}^{2} - {t}^{2}\right)\left(1 + {e}^{- 4t}\right)} + \dfrac{1}{\left({\left(\dfrac{\sqrt{11}}{2}\right)}^{2} - {t}^{2}\right)\left(1 + {e}^{4t}\right)}dt$

$⇒ I = \dfrac{1}{2}{∫}_{0}^{\dfrac{1}{2}}\dfrac{{e}^{4t}}{\left({\left(\dfrac{\sqrt{11}}{2}\right)}^{2} - {t}^{2}\right)\left({e}^{4t} + 1\right)} + \dfrac{1}{\left({\left(\dfrac{\sqrt{11}}{2}\right)}^{2} - {t}^{2}\right)\left(1 + {e}^{4t}\right)}dt$

$⇒ I = \dfrac{1}{2}{∫}_{0}^{\dfrac{1}{2}}\dfrac{1}{{\left(\dfrac{\sqrt{11}}{2}\right)}^{2} - {t}^{2}}dt$

$⇒ I = \dfrac{1}{2} \times \dfrac{1}{2\left(\dfrac{\sqrt{11}}{2}\right)}{\left[\ln \left|\dfrac{\dfrac{\sqrt{11}}{2} + t}{\dfrac{\sqrt{11}}{2} - t}\right|\right]}_{0}^{\dfrac{1}{2}}$

$⇒ I = \dfrac{1}{2\sqrt{11}}\ln \left(\dfrac{\sqrt{11} + 1}{\sqrt{11} - 1}\right) = \dfrac{1}{2\sqrt{11}}\ln \left(\dfrac{( \sqrt{11} + 1{)}^{2}}{10}\right)$

$⇒ I = \dfrac{1}{\sqrt{11}}\ln \left(\dfrac{\sqrt{11} + 1}{\sqrt{10}}\right)$

Now on comparing with $\dfrac{1}{\alpha}\log_{e}\left(\dfrac{\alpha + 1}{\beta}\right)$ we get,

$⇒ \alpha = \sqrt{11}, \beta = \sqrt{10} ⇒ {\alpha}^{4} - {\beta}^{4} = 21$