Let f(x) = integral (16x + 24)/(x^2 + 2x - 15) dx. If f(4) = 14 log_e(3) and f(7) = log_e(2^alpha * 3^beta), alpha, beta in N, then alpha + beta is equal to :

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q19 – JEE-MAIN 2026

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Home Maths Limits and DerivativesQuestion #253

Let

f (x) = \int \frac{16 x + 24}{x^{2} + 2 x - 15} d x

. If

f (4) = 14 \log_{e} (3)

and

f (7) = \log_{e} (2^{α} \cdot 3^{β})

α, β \in N

, then

α + β

is equal to :

Clue 1 of 3: Integral Evaluation

The first step is to evaluate the indefinite integral $f (x)$ using appropriate integration techniques.

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Step 1: Evaluate the indefinite integral

f (x)

✦ Active

First, factor the denominator: $x^{2} + 2 x - 15 = (x + 5) (x - 3)$ . Then, use partial fraction decomposition for the integrand:

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

Solving for $A$ and $B$ : Set $x = - 5 \Rightarrow 16 (- 5) + 24 = A (- 5 - 3) \Rightarrow - 56 = - 8 A \Rightarrow A = 7$ . Set $x = 3 \Rightarrow 16 (3) + 24 = B (3 + 5) \Rightarrow 72 = 8 B \Rightarrow B = 9$ . Thus, the integral becomes:

f (x) = \int (\frac{7}{x + 5} + \frac{9}{x - 3}) d x = 7 \log_{e} | x + 5 | + 9 \log_{e} | x - 3 | + C

Step 2: Determine the constant of integration

C

○ Expand

Use the given condition $f (4) = 14 \log_{e} (3)$ to find $C$ . Since $x = 4$ , $x + 5$ and $x - 3$ are positive, so we can remove the absolute values.

f (4) = 7 \log_{e} (4 + 5) + 9 \log_{e} (4 - 3) + C

$f (4) = 7 \log_{e} (9) + 9 \log_{e} (1) + C = 7 \log_{e} (3^{2}) + 0 + C = 14 \log_{e} (3) + C$

Equating this with the given value: $14 \log_{e} (3) = 14 \log_{e} (3) + C$ , which implies $C = 0$ . Therefore, for $x > 3$ , $f (x) = 7 \log_{e} (x + 5) + 9 \log_{e} (x - 3)$ .

💡 Teacher's Secret Hint

Ensure to correctly handle the constant of integration using the initial condition.

Step 3: Calculate

f (7)

and find

α + β

○ Expand

Substitute $x = 7$ into the determined $f (x)$ :

f (7) = 7 \log_{e} (7 + 5) + 9 \log_{e} (7 - 3) = 7 \log_{e} (12) + 9 \log_{e} (4)

Express $12$ and $4$ in terms of prime factors ( $12 = 2^{2} \cdot 3$ , $4 = 2^{2}$ ) and use logarithm properties:

f (7) = 7 \log_{e} (2^{2} \cdot 3) + 9 \log_{e} (2^{2})

f (7) = 7 (2 \log_{e} (2) + \log_{e} (3)) + 9 (2 \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)

We are given $f (7) = \log_{e} (2^{α} \cdot 3^{β}) = α \log_{e} (2) + β \log_{e} (3)$ . Comparing the coefficients with our result:

α = 32 and β = 7

Both $α$ and $β$ are natural numbers. The required sum is $α + β = 32 + 7 = 39$ .

💡 Teacher's Secret Hint

Carefully apply logarithm properties to simplify the expression for $f (7)$ and match the coefficients.

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Maths Question 19 – JEE-MAIN 2026

Clue 1 of 3: Integral Evaluation

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