Question

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 \( \alpha, \beta \in \mathbb{N} \), then \( \alpha + \beta \) is:

• A. 31
• B. 37
• C. 39
• D. 41

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We are given an indefinite integral of a rational function.
The denominator can be factored, meaning we can use the method of partial fractions to integrate it.
After finding the general antiderivative \(f(x)\), we will use the initial condition \(f(4)\) to find the constant of integration \(C\).
Finally, we evaluate \(f(7)\) and match it to the given logarithmic form to find \(\alpha\) and \(\beta\).
Step 2: Key Formula or Approach:
Factor the denominator: \(x^2 + 2x - 15 = (x + 5)(x - 3)\).
Set up the partial fraction decomposition.
\[ \frac{16x + 24}{(x + 5)(x - 3)} = \frac{A}{x + 5} + \frac{B}{x - 3} \] Use the cover-up method to find constants \(A\) and \(B\).
Step 3: Detailed Explanation:
To find \(A\), cover \((x+5)\) and substitute \(x = -5\).
\[ A = \frac{16(-5) + 24}{-5 - 3} = \frac{-80 + 24}{-8} = \frac{-56}{-8} = 7 \] To find \(B\), cover \((x-3)\) and substitute \(x = 3\).
\[ B = \frac{16(3) + 24}{3 + 5} = \frac{48 + 24}{8} = \frac{72}{8} = 9 \] Now integrate the separated fractions.
\[ f(x) = \int \left( \frac{7}{x + 5} + \frac{9}{x - 3} \right) dx = 7\ln|x + 5| + 9\ln|x - 3| + C \] Use the condition \(f(4) = 14\ln(3)\) to find \(C\).
\[ f(4) = 7\ln|4 + 5| + 9\ln|4 - 3| + C \] \[ f(4) = 7\ln(9) + 9\ln(1) + C \] Since \(\ln(9) = \ln(3^2) = 2\ln(3)\) and \(\ln(1) = 0\):
\[ f(4) = 7(2\ln 3) + 0 + C = 14\ln 3 + C \] Given \(f(4) = 14\ln 3\), we conclude that \(C = 0\).
So the function is \(f(x) = 7\ln|x + 5| + 9\ln|x - 3|\).
Now, evaluate \(f(7)\).
\[ f(7) = 7\ln(12) + 9\ln(4) \] Express everything in terms of prime bases 2 and 3.
\[ \ln(12) = \ln(2^2 \cdot 3) = 2\ln 2 + \ln 3 \] \[ \ln(4) = \ln(2^2) = 2\ln 2 \] Substitute these back.
\[ f(7) = 7(2\ln 2 + \ln 3) + 9(2\ln 2) \] \[ f(7) = 14\ln 2 + 7\ln 3 + 18\ln 2 = 32\ln 2 + 7\ln 3 \] Combine into a single logarithm using logarithm power rules.
\[ f(7) = \ln(2^{32}) + \ln(3^7) = \ln(2^{32} \cdot 3^7) \] Comparing this to \(\ln(2^\alpha \cdot 3^\beta)\), we get \(\alpha = 32\) and \(\beta = 7\).
The requested sum is \(\alpha + \beta = 32 + 7 = 39\).
Step 4: Final Answer:
The value of \(\alpha + \beta\) is \(39\).