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

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

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

$$ 2I=\int_0^1 \frac{dx}{2\left(\frac{11}{4}-\left(x-\frac{1}{2}\right)^2\right)} $$

To solve this integral, we can perform a change of variables using the substitution

$x-\frac{1}{2}= \frac{\sqrt{11}}{2}\tan{u}$, then $dx=\frac{\sqrt{11}}{2}\sec^2{u} du$:

$$ I=\frac{1}{\sqrt{11}}\int \frac{\sec^2{u}}{1+\tan^2{u}}du $$

$$ I=\frac{1}{\sqrt{11}}\int du = \frac{1}{\sqrt{11}}(u+C) $$

Now we need to find the limits of the integral after the substitution. If $x=0$, then $u=\tan^{-1}\left(-\frac{1}{\sqrt{11}}\right)$. If $x=1$, then $u=\tan^{-1}\left(\frac{1}{\sqrt{11}}\right)$. So, the integral becomes:

$$ I=\frac{1}{\sqrt{11}}\left[\tan^{-1}\left(\frac{1}{\sqrt{11}}\right)-\tan^{-1}\left(-\frac{1}{\sqrt{11}}\right)\right] $$

Using the properties of the arctangent function, we can rewrite the integral as:

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

From this result, we have $\alpha=\sqrt{11}$ and $\beta=\sqrt{10}$. Now, we can find $\alpha^4 - \beta^4$:

$$ \alpha^4 - \beta^4 = (11)^2 - (10)^2 = 121 - 100 = 21 $$

Thus, $\alpha^4 - \beta^4 = 21$.

Definite Integration

Continue this paper