Let f(x) = ∫ ((16x + 24)/(x^2 + 2x - 15)) dx . If f(4) = 14 _e(3)...

The given integral is $f(x) = \int \left(\dfrac{16x + 24}{x^2 + 2x - 15}\right) dx$.

Factorizing the denominator, we get $x^2 + 2x - 15 = (x + 5)(x - 3)$.

Using partial fractions, we can write:

$\dfrac{16x + 24}{(x + 5)(x - 3)} = \dfrac{A}{x + 5} + \dfrac{B}{x - 3}$

$16x + 24 = A(x - 3) + B(x + 5)$

Substituting $x = 3$, we get $72 = 8B \Rightarrow B = 9$.

Substituting $x = -5$, we get $-56 = -8A \Rightarrow A = 7$.

Therefore, the integral becomes:

$f(x) = \int \left(\dfrac{7}{x + 5} + \dfrac{9}{x - 3}\right) dx$

$f(x) = 7 \log_e|x + 5| + 9 \log_e|x - 3| + C$

Given $f(4) = 14 \log_e(3)$, we substitute $x = 4$:

$f(4) = 7 \log_e(9) + 9 \log_e(1) + C = 14 \log_e(3) + C$

Since $f(4) = 14 \log_e(3)$, we get $C = 0$.

Thus, $f(x) = 7 \log_e|x + 5| + 9 \log_e|x - 3|$.

Now, substituting $x = 7$:

$f(7) = 7 \log_e(12) + 9 \log_e(4)$

$f(7) = 7 \log_e(2^2 \cdot 3) + 9 \log_e(2^2)$

$f(7) = 7(2 \log_e(2) + \log_e(3)) + 18 \log_e(2)$

$f(7) = 14 \log_e(2) + 7 \log_e(3) + 18 \log_e(2)$

$f(7) = 32 \log_e(2) + 7 \log_e(3)$

$f(7) = \log_e(2^{32} \cdot 3^7)$

Comparing this with $f(7) = \log_e(2^{\alpha} \cdot 3^{\beta})$, we get $\alpha = 32$ and $\beta = 7$.

Therefore, $\alpha + \beta = 32 + 7 = 39$.

Answer: $39$