JEE Mains · Maths · STD 12 - 7.1 indefinite integral
Let \(f(x) = \int \left(\dfrac{16x + 24}{x^2 + 2x - 15}\right) dx\). If \(f(4) = 14 \log_e(3)\) and \(f(7) = \log_e(2^{\alpha} \cdot 3^{\beta})\), \(\alpha, \beta \in \mathbb{N}\), then \(\alpha + \beta\) is equal to:
- A \(31\)
- B \(37\)
- C \(39\)
- D \(41\)
Answer & Solution
Correct Answer
(C) \(39\)
Step-by-step Solution
Detailed explanation
The given integral is \(f(x) = \int \left(\dfrac{16x + 24}{x^2 + 2x - 15}\right) dx\). Factorizing the denominator, we get \(x^2 + 2x - 15 = (x + 5)(x - 3)\). Using partial fractions, we can write: \(\dfrac{16x + 24}{(x + 5)(x - 3)} = \dfrac{A}{x + 5} + \dfrac{B}{x - 3}\)…
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