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Step-by-Step Solution
Step 1: Write down the given integral
We want to evaluate
$I = \int_{0}^{1} \frac{1}{\bigl(5 + 2x - 2x^2\bigr)\bigl(1 + e^{2 - 4x}\bigr)} \, dx$.
It is also given that
$I = \frac{1}{\alpha}\,\ln\bigl(\tfrac{\alpha + 1}{\beta}\bigr)$,
and we wish to find
$\alpha^{4} - \beta^{4}$.
Step 2: Consider the substitution $x \mapsto 1 - x$
Let $u = 1 - x$. Then $dx = -\,du$. When $x = 0$, $u = 1$; and when $x = 1$, $u = 0$. So
$
I = \int_{1}^{0} \frac{-\,du}{\bigl(5 + 2(1 - u) - 2(1 - u)^2\bigr)\,\bigl(1 + e^{2 - 4(1 - u)}\bigr)}.
$
Reversing the limits (from 1 to 0, to 0 to 1) introduces a factor of $-1$ that cancels the $-\,du$. Therefore,
$
I = \int_{0}^{1} \frac{du}{\bigl(5 + 2(1 - u) - 2(1 - u)^2\bigr)\,\bigl(1 + e^{2 - 4(1 - u)}\bigr)}.
$
Step 3: Simplify the integrand after substitution
Inside the quadratic term:
$5 + 2(1 - u) - 2(1 - u)^2
= 5 + 2 - 2u - 2\bigl(1 - 2u + u^2\bigr)
= 7 - 2u -2 + 4u - 2u^2
= 5 + 2u - 2u^2.$
Meanwhile,
$1 + e^{2 - 4(1 - u)}
= 1 + e^{2 - 4 + 4u}
= 1 + e^{4u - 2}.$
Notice this is very similar to the original integrand, merely swapping $e^{2 - 4x}$ with $e^{4x - 2}$ when $u$ replaces $x$.
Step 4: Add the original and transformed integrals
Let $I_1$ be the original integral in terms of $x$ and $I_2$ be the newly expressed integral in terms of $u$. By symmetry, $I_1 = I_2 = I$. When we add them together, the denominators combine in such a way that
$
I_1 + I_2
= \int_{0}^{1} \frac{1}{5 + 2x - 2x^2} \,dx.
$
Since $I_1 = I_2 = I$, it follows
$
2I = \int_{0}^{1} \frac{dx}{5 + 2x - 2x^2}.
$
Hence,
$
I = \frac{1}{2} \int_{0}^{1} \frac{dx}{5 + 2x - 2x^2}.
$
Step 5: Complete the square for $5 + 2x - 2x^2$
First rewrite:
$5 + 2x - 2x^2 = -\,2\bigl(x^2 - x - \tfrac{5}{2}\bigr).$
Complete the square inside the parentheses:
$
x^2 - x - \tfrac{5}{2}
= x^2 - x + \bigl(\tfrac{1}{2}\bigr)^2 - \bigl(\tfrac{1}{2}\bigr)^2 - \tfrac{5}{2}
= \bigl(x - \tfrac{1}{2}\bigr)^2 - \tfrac{1}{4} - \tfrac{5}{2}
= \bigl(x - \tfrac{1}{2}\bigr)^2 - \tfrac{11}{4}.
$
Therefore,
$
5 + 2x - 2x^2
= -\,2 \Bigl[\bigl(x - \tfrac{1}{2}\bigr)^2 - \tfrac{11}{4}\Bigr]
= 2 \Bigl[\tfrac{11}{4} - \bigl(x - \tfrac{1}{2}\bigr)^2\Bigr].
$
Step 6: Rewrite the integral in a more standard form
Using this factorization, we have
$
\int_{0}^{1} \frac{dx}{5 + 2x - 2x^2}
= \int_{0}^{1} \frac{dx}{2\Bigl[\tfrac{11}{4} - (x - \tfrac{1}{2})^2\Bigr]}.
$
Hence
$
2I
= \int_{0}^{1} \frac{dx}{2\Bigl[\tfrac{11}{4} - (x - \tfrac{1}{2})^2\Bigr]}
= \frac{1}{2} \int_{0}^{1} \frac{dx}{\tfrac{11}{4} - (x - \tfrac{1}{2})^2}.
$
Therefore,
$
I = \frac{1}{4} \int_{0}^{1} \frac{dx}{\tfrac{11}{4} - (x - \tfrac{1}{2})^2}.
$
Step 7: Use the standard integral involving a quadratic in the denominator
Recall the known result:
$
\int \frac{dx}{a^2 - (x - h)^2}
= \frac{1}{2a} \,\ln \Bigl| \frac{a + (x - h)}{a - (x - h)} \Bigr| + C,
$
where $a = \tfrac{\sqrt{11}}{2}$ in our case and $h = \tfrac{1}{2}$. Evaluating from $x=0$ to $x=1$ then gives a logarithmic expression. Matching it with
$
I = \frac{1}{\alpha}\,\ln\Bigl(\frac{\alpha + 1}{\beta}\Bigr)
$
shows that
$
\alpha = \sqrt{11}, \quad \beta = \sqrt{10}.
$
Step 8: Compute $\alpha^4 - \beta^4$
Finally,
$
\alpha^4 - \beta^4
= (\sqrt{11})^4 - (\sqrt{10})^4
= 11^2 - 10^2
= 121 - 100
= 21.
$
Answer: 21
