Question

If \( I(x) = \int \frac{16x+24}{x^2+2x-15}\,dx \), \( I(1) = 14\ln 3 \) and \( I(7) = \ln\left(2^\alpha \cdot 3^\beta\right) \), then \( (\alpha+\beta) \) is equal to

• A. 39
• B. 33
• C. 36
• D. 42

Hint:

When an integral contains a rational function with a factorable quadratic denominator, partial fraction decomposition is usually the fastest method. After integrating, use the given condition carefully to determine the constant of integration.

Answer 1

The Correct Option is A

Step 1: Factorize the denominator and decompose into partial fractions.
We have:
\[ x^2+2x-15=(x+5)(x-3) \] So,
\[ \frac{16x+24}{x^2+2x-15}=\frac{16x+24}{(x+5)(x-3)} \] Let:
\[ \frac{16x+24}{(x+5)(x-3)}=\frac{A}{x+5}+\frac{B}{x-3} \] Then:
\[ 16x+24=A(x-3)+B(x+5) \] \[ 16x+24=(A+B)x+(-3A+5B) \] Comparing coefficients, we get:
\[ A+B=16 \] \[ -3A+5B=24 \] Solving these,
\[ A=7,\qquad B=9 \] Hence,
\[ \frac{16x+24}{x^2+2x-15}=\frac{7}{x+5}+\frac{9}{x-3} \]
Step 2: Integrate to find \(I(x)\).
Therefore,
\[ I(x)=\int \left( \frac{7}{x+5}+\frac{9}{x-3} \right)\,dx \] \[ I(x)=7\ln|x+5|+9\ln|x-3|+C \]
Step 3: Use the condition \(I(D)=14\ln 3\) to find \(C\).
Substituting \(x=4\),
\[ I(D)=7\ln|4+5|+9\ln|4-3|+C \] \[ I(D)=7\ln 9+9\ln 1+C \] Since \( \ln 1=0 \),
\[ I(D)=7\ln 9+C \] \[ I(D)=7\cdot 2\ln 3+C=14\ln 3+C \] But it is given that \(I(D)=14\ln 3\). Hence,
\[ C=0 \] So,
\[ I(x)=7\ln|x+5|+9\ln|x-3| \]
Step 4: Find \(I(7)\).
Now substitute \(x=7\):
\[ I(7)=7\ln|7+5|+9\ln|7-3| \] \[ I(7)=7\ln 12+9\ln 4 \] Now,
\[ \ln 12=\ln(2^2\cdot 3)=2\ln 2+\ln 3 \] and
\[ \ln 4=2\ln 2 \] So,
\[ I(7)=7(2\ln 2+\ln 3)+9(2\ln 2) \] \[ I(7)=14\ln 2+7\ln 3+18\ln 2 \] \[ I(7)=32\ln 2+7\ln 3 \] \[ I(7)=\ln(2^{32})+\ln(3^7)=\ln(2^{32}\cdot 3^7) \] Thus,
\[ \alpha=32,\qquad \beta=7 \]
Step 5: Find \( \alpha+\beta \).
\[ \alpha+\beta=32+7=39 \]
Step 6: Conclusion.
Hence, the required value of \( \alpha+\beta \) is \(39\).

Final Answer: \(39\)