If $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{25 x^2 \cos^2 x}{2(x + e)^2} \, dx = \pi (\alpha \pi^2 + \beta)$, $\alpha, \beta \in \mathbb{Z}$, then $(\alpha + \beta)^2$ equals

Step 1: Analyze the integrand and limits

The given integral has symmetric limits from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$. This suggests using properties of definite integrals for even and odd functions. However, the integrand is not a simple even or odd function due to the $(x+e)^2$ term.

Step 2: Apply the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$

Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$, where $a = -\frac{\pi}{2}$ and $b = \frac{\pi}{2}$, we have $a+b-x = -x$. Substituting this into the integrand, we get a new expression for $I$. Since $\cos(-x) = \cos x$ and $(-x)^2 = x^2$, the numerator remains $25x^2 \cos^2 x$.

Step 3: Combine the original and transformed integrals

Let the original integral be $I_1$ and the transformed integral be $I_2$. Since $I_1 = I_2 = I$, we can add them: $2I = I_1 + I_2$. This allows us to combine the denominators. We factor out the common term $25x^2 \cos^2 x$.

Step 4: Simplify the combined fraction

We simplify the sum of the fractions by finding a common denominator. Expanding the terms in the numerator and denominator, we notice that the $2xe$ terms cancel out, leading to a simplified expression. This simplification is crucial for further steps.

Step 5: Re-evaluate the problem statement

Upon simplifying the sum of the two integrals, we arrive at an integrand that is still complex and does not appear to simplify further into a form that would yield the expected result $\pi (\alpha \pi^2 + \beta)$ with integer $\alpha$ and $\beta$. This suggests that the problem statement might contain a typo, possibly in the denominator or the constant factor, as the current form is not standard for competitive exams or typical integral properties.

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