CUET · MATHS · PYQ PAPER 2025
\(\int \frac{d x}{9 x^2-16}\) is equal to:
- A \(\frac{1}{24} \log _e\left|\frac{3 x+4}{3 x-4}\right|+C\), Where \(C\) is constant of integration
- B \(\frac{3}{8} \log _e\left|\frac{3 x+4}{3 x-4}\right|+C\), Where \(C\) is constant of integration
- C \(\frac{3}{8} \log _e\left|\frac{3 x-4}{3 x+4}\right|+C\), Where \(C\) is constant of integration
- D \(\frac{1}{24} \log _e\left|\frac{3 x-4}{3 x+4}\right|+C\), Where \(C\) is constant of integration
Answer & Solution
Correct Answer
(D) \(\frac{1}{24} \log _e\left|\frac{3 x-4}{3 x+4}\right|+C\), Where \(C\) is constant of integration
Step-by-step Solution
Detailed explanation
\( \int \frac{d x}{9 x^2-16} = \int \frac{d x}{(3x)^2-4^2} \) Let \( u = 3x \implies du = 3dx \implies dx = \frac{1}{3} du \) \( = \frac{1}{3} \int \frac{du}{u^2-4^2} \) \( = \frac{1}{3} \left( \frac{1}{2 \times 4} \log_e\left|\frac{u-4}{u+4}\right| \right) + C \)…
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