Let f(x) = int frac{16x + 24}{x^2 + 2x - 15} | vedclass

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(C) First, we decompose the integrand into partial fractions:$\frac{16x+24}{(x+5)(x-3)} = \frac{A}{x+5} + \frac{B}{x-3}$$16x+24 = A(x-3) + B(x+5)$Sett

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$39$

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(C) First, we decompose the integrand into partial fractions:$\frac{16x+24}{(x+5)(x-3)} = \frac{A}{x+5} + \frac{B}{x-3}$$16x+24 = A(x-3) + B(x+5)$Setting $x=3$, we get $72 = 8B \implies B=9$.Setting $x=-5$, we get $-56 = -8A \implies A=7$.Thus, $f(x) = \int (\frac{7}{x+5} + \frac{9}{x-3}) dx = 7 \log|x+5| + 9 \log|x-3| + C$.Given $f(4) = 14 \log 3$, we have $7 \log 9 + 9 \log 1 + C = 14 \log 3 + C = 14 \log 3$, which implies $C=0$.Now, calculate $f(7) = 7 \log|7+5| + 9 \log|7-3| = 7 \log 12 + 9 \log 4$.$f(7) = 7 \log(2^2 \cdot 3) + 9 \log(2^2) = 7(2 \log 2 + \log 3) + 18 \log 2 = 14 \log 2 + 7 \log 3 + 18 \log 2 = 32 \log 2 + 7 \log 3 = \log(2^{32} \cdot 3^7)$.Comparing with $\log(2^\alpha \cdot 3^\beta)$, we get $\alpha = 32$ and $\beta = 7$.Therefore, $\alpha + \beta = 32 + 7 = 39$.

Let $f(x) = \int \frac{16x + 24}{x^2 + 2x - 15} dx$. If $f(4) = 14 \log_e(3)$ and $f(7) = \log_e(2^\alpha \cdot 3^\beta)$, where $\alpha, \beta \in N$, then $\alpha + \beta$ is equal to:

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