Find Sum of Exponents in Logarithmic Value of Indefinite Integral

To evaluate the given indefinite integral, we first decompose the integrand into partial fractions.

Given: $f(x) = \int \left( \frac{16x + 24}{x^2 + 2x - 15} \right) \mathrm{d}x$

First, factor the denominator: $x^2 + 2x - 15 = (x + 5)(x - 3)$

Now, express the integrand in terms of partial fractions: $\frac{16x + 24}{(x+5)(x-3)} = \frac{A}{x+5} + \frac{B}{x-3}$

Multiplying both sides by $(x+5)(x-3)$, we get: $16x + 24 = A(x - 3) + B(x + 5)$

To find $B$, substitute $x = 3$: $16(3) + 24 = B(3 + 5)$ $72 = 8B \implies B = 9$

To find $A$, substitute $x = -5$: $16(-5) + 24 = A(-5 - 3)$ $-56 = -8A \implies A = 7$

Thus, the integrand becomes: $\frac{16x + 24}{x^2 + 2x - 15} = \frac{7}{x+5} + \frac{9}{x-3}$

Integrating both sides with respect to $x$: $f(x) = \int \left( \frac{7}{x+5} + \frac{9}{x-3} \right) \mathrm{d}x = 7 \log_e|x+5| + 9 \log_e|x-3| + C$

We are given $f(4) = 14 \log_e(3)$. Substitute $x = 4$ into $f(x)$: $f(4) = 7 \log_e|4+5| + 9 \log_e|4-3| + C$ $f(4) = 7 \log_e(9) + 9 \log_e(1) + C$ $f(4) = 7 \log_e(3^2) + 0 + C$ $f(4) = 14 \log_e(3) + C$

Given $f(4) = 14 \log_e(3)$, we get: $14 \log_e(3) + C = 14 \log_e(3) \implies C = 0$

Therefore, the function is: $f(x) = 7 \log_e|x+5| + 9 \log_e|x-3|$

Now, calculate $f(7)$: $f(7) = 7 \log_e|7+5| + 9 \log_e|7-3|$ $f(7) = 7 \log_e(12) + 9 \log_e(4)$

Using the prime factorizations $12 = 2^2 \cdot 3$ and $4 = 2^2$: $f(7) = 7 \log_e(2^2 \cdot 3) + 9 \log_e(2^2)$ $f(7) = 7 (2 \log_e 2 + \log_e 3) + 9 (2 \log_e 2)$ $f(7) = 14 \log_e 2 + 7 \log_e 3 + 18 \log_e 2$ $f(7) = 32 \log_e 2 + 7 \log_e 3$ $f(7) = \log_e(2^{32}) + \log_e(3^7)$ $f(7) = \log_e(2^{32} \cdot 3^7)$

Comparing this with the given expression $f(7) = \log_e(2^\alpha \cdot 3^\beta)$, where $\alpha, \beta \in \mathbb{N}$: $\alpha = 32$ $\beta = 7$

Hence, the value of $\alpha + \beta$ is: $\alpha + \beta = 32 + 7 = 39$

Correct Option: C