If int limits_0^1 frac{1}{left(5+2 x -2 x ^2r | vedclass

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(A) Let $I = \int \limits_0^1 \frac{dx}{(5+2x-2x^2)(1+e^{2-4x})}$.Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we replace $x$ with $1-x$

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$21$

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(A) Let $I = \int \limits_0^1 \frac{dx}{(5+2x-2x^2)(1+e^{2-4x})}$.Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we replace $x$ with $1-x$:$I = \int \limits_0^1 \frac{dx}{(5+2(1-x)-2(1-x)^2)(1+e^{2-4(1-x)})} = \int \limits_0^1 \frac{dx}{(5+2-2x-2(1-2x+x^2))(1+e^{4x-2})}$$I = \int \limits_0^1 \frac{dx}{(5+2-2x-2+4x-2x^2)(1+e^{-(2-4x)})} = \int \limits_0^1 \frac{e^{2-4x} dx}{(5+2x-2x^2)(1+e^{2-4x})}$.Adding the two expressions for $I$:$2I = \int \limits_0^1 \frac{1+e^{2-4x}}{(5+2x-2x^2)(1+e^{2-4x})} dx = \int \limits_0^1 \frac{dx}{5+2x-2x^2}$.Completing the square: $5+2x-2x^2 = -2(x^2-x) + 5 = -2(x^2-x+\frac{1}{4}) + 5 + \frac{1}{2} = \frac{11}{2} - 2(x-\frac{1}{2})^2 = 2(\frac{11}{4} - (x-\frac{1}{2})^2)$.$2I = \frac{1}{2} \int_0^1 \frac{dx}{(\frac{\sqrt{11}}{2})^2 - (x-\frac{1}{2})^2} = \frac{1}{2} \cdot \frac{1}{2(\frac{\sqrt{11}}{2})} \ln \left| \frac{\frac{\sqrt{11}}{2} + (x-\frac{1}{2})}{\frac{\sqrt{11}}{2} - (x-\frac{1}{2})} \right|_0^1 = \frac{1}{2\sqrt{11}} \ln \left( \frac{\sqrt{11}+1}{\sqrt{11}-1} \right) = \frac{1}{\sqrt{11}} \ln \left( \frac{\sqrt{11}+1}{\sqrt{10}} \right)$ (after rationalizing).Comparing with $\frac{1}{\alpha} \ln \left( \frac{\alpha+1}{\beta} \right)$,we get $\alpha = \sqrt{11}$ and $\beta = \sqrt{10}$.Thus,$\alpha^4 - \beta^4 = (\sqrt{11})^4 - (\sqrt{10})^4 = 121 - 100 = 21$.

If $\int \limits_0^1 \frac{1}{\left(5+2 x -2 x ^2\right)\left(1+ e ^{(2-4 x)}\right)} dx =\frac{1}{\alpha} \log _{ e }\left(\frac{\alpha+1}{\beta}\right)$ where $\alpha, \beta > 0$,then $\alpha^4-\beta^4$ is equal to:

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