Let $f(x) = \int\left(\frac{16x + 24}{x^2 + 2x - 15}\right)dx$. If $f(4) = 14 \log_e(3)$ and $f(7) = \log_e(2^{\alpha} \cdot 3^{\beta})$, $\alpha, \beta \in N$, then $\alpha + \beta$ is equal to :
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Answer: 39
Key idea: Partial Fractions and Logarithmic Integration
- Step 1: Factor the denominator: $x^2 + 2x - 15 = (x+5)(x-3)$
- Step 2: Use partial fractions: $\frac{16x + 24}{(x+5)(x-3)} = \frac{7}{x+5} + \frac{9}{x-3}$ (by solving $16x + 24 = A(x-3) + B(x+5)$)
- Step 3: Integrate to get $f(x) = 7\ln|x+5| + 9\ln|x-3| + C$
- Step 4: Use $f(4) = 14\ln(3)$ to find $C$: $7\ln(9) + 9\ln(1) + C = 14\ln(3)$, which gives $C = 0$
- Step 5: Calculate $f(7) = 7\ln(12) + 9\ln(4) = 7\ln(2^2 \cdot 3) + 9\ln(2^2) = 32\ln(2) + 7\ln(3) = \ln(2^{32} \cdot 3^7)$
- Step 6: Compare with $\ln(2^{\alpha} \cdot 3^{\beta})$ to get $\alpha = 32$, $\beta = 7$, so $\alpha + \beta = 39$