If $$\int_0^1 \frac{1}{(5+2x-2x^2)(1+e^{(2-4x)})} dx = \frac{1}{\alpha} \log_e\left(\frac{\alpha+1}{\beta}\right)$$, $$\alpha, \beta > 0$$, then $$\alpha^4 - \beta^4$$ is equal to

$$I = \int_0^1 \frac{1}{(5 + 2x - 2x^2)(1 + e^{2-4x})} dx$$

Applying property $$\int_0^a f(x) dx = \int_0^a f(a-x) dx$$:

$$I = \int_0^1 \frac{1}{(5 + 2(1-x) - 2(1-x)^2)(1 + e^{2-4(1-x)})} dx$$

$$I = \int_0^1 \frac{1}{(5 + 2 - 2x - 2(1 - 2x + x^2))(1 + e^{4x-2})} dx$$

$$I = \int_0^1 \frac{1}{(5 + 2x - 2x^2)(1 + e^{-(2-4x)})} dx = \int_0^1 \frac{e^{2-4x}}{(5 + 2x - 2x^2)(e^{2-4x} + 1)} dx$$

$$2I = \int_0^1 \frac{1 + e^{2-4x}}{(5 + 2x - 2x^2)(1 + e^{2-4x})} dx = \int_0^1 \frac{1}{5 + 2x - 2x^2} dx$$

$$2I = \int_0^1 \frac{1}{5 - 2(x^2 - x)} dx = \int_0^1 \frac{1}{5 - 2\left(x^2 - x + \frac{1}{4} - \frac{1}{4}\right)} dx$$

$$2I = \int_0^1 \frac{1}{5 + \frac{1}{2} - 2\left(x - \frac{1}{2}\right)^2} dx = \int_0^1 \frac{1}{\frac{11}{2} - 2\left(x - \frac{1}{2}\right)^2} dx$$

$$2I = \frac{1}{2} \int_0^1 \frac{1}{\frac{11}{4} - \left(x - \frac{1}{2}\right)^2} dx = \frac{1}{2} \left[ \frac{1}{2\left(\frac{\sqrt{11}}{2}\right)} \log_e \left| \frac{\frac{\sqrt{11}}{2} + x - \frac{1}{2}}{\frac{\sqrt{11}}{2} - x + \frac{1}{2}} \right| \right]_0^1$$

$$I = \frac{1}{2\sqrt{11}} \log_e \left( \frac{\sqrt{11}+1}{\sqrt{11}-1} \right) = \frac{1}{2\sqrt{11}} \log_e \left( \frac{(\sqrt{11}+1)^2}{11-1} \right) = \frac{1}{\sqrt{11}} \log_e \left( \frac{\sqrt{11}+1}{\sqrt{10}} \right)$$

Comparing with $$\frac{1}{\alpha} \log_e \left( \frac{\alpha+1}{\beta} \right)$$:

$$\alpha = \sqrt{11}, \beta = \sqrt{10}$$

$$\alpha^4 - \beta^4 = (\sqrt{11})^4 - (\sqrt{10})^4 = 121 - 100 = 21$$