Let f ( x ) = int (frac 16 x + 24 x 2 + 2 x - 15 ) d x . If f (4) = 14log e(3) and f (7) = log e(2 α .3 β ) , α ,β in N , then α +β is equal to:

Mathematics · Indefinite Integration

JEE Main 2026 — 2 April, Evening Shift — Question 41

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

Option A:
31
Option B:
37
Option C:
39
Correct
Option D:
41

Answer: C

Step-by-step solution

$f(x)=\int \frac{8(2 x+2)+8}{x^{2}+2 x-15} d x$ $\Rightarrow \mathrm{f}(\mathrm{x})=8 \ell \mathrm{n}\left|\mathrm{x}^{2}+2 \mathrm{x}-15\right|+\ell \mathrm{n}\left|\frac{\mathrm{x}-3}{\mathrm{x}+5}\right|+\mathrm{C}$ $\Rightarrow \mathrm{f}(4)=14 \ell \mathrm{n} 3+\mathrm{C}$ $\Rightarrow 14 \ell \mathrm{n} 3=14 \ell \mathrm{n} 3+\mathrm{C}$ $\therefore \mathrm{C}=0$ Now, $f(7)=8 \ell \mathrm{n} 48-\ell \mathrm{n} 3$ $=\ell n\left(\frac{(48)^{8}}{3}\right)$ $\Rightarrow \mathrm{f}(7)=\ln \left(2^{32} \times 3^{7}\right)$ $\alpha=32, \beta=7 \Rightarrow \alpha+\beta=39$

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Exam: JEE Main 2026
Paper: 2 April, Evening Shift
Subject: Mathematics
Chapter: Indefinite Integration
Topic: Methods of Indefinite Integration