If $\underset{0}{\overset{1}{\int }}\frac{1}{(5+2x-2x^2)(1+e^{(2-4x)})}dx=\frac{1}{\alpha }{\log }_e\left(\frac{\alpha +1}{\beta }\right), \alpha ,\beta >0,$ then ${\alpha }^4-{\beta }^4-({\alpha }^2+{\beta }^2)$ is equal to  

Let, $I=\underset{0}{\overset{1}{\int }}\frac{dx}{(5+2x-2x^2)(1+e^{2-4x})}$

Apply King’s Rule $I=\underset{0}{\overset{1}{\int }}\frac{e^{2-4x}dx}{(5+2x-2x^2)(1+e^{2-4x})}$

$2I=\int \frac{1dx}{5+2x-2x^2}=\underset{0}{\overset{1}{\int }}\frac{dx}{2(\frac{11}{4}-{(x-\frac{1}{2})}^2)}$

$I=\frac{1}{\sqrt{11}}\ln \left(\frac{\sqrt{11}+1}{\sqrt{10}}\right)\Rightarrow \alpha =\sqrt{11},\beta =\sqrt{10}$

so, ${\alpha }^4-{\beta }^4=121-100=21$