CUET · MATHS · PYQ PAPER 2023
\(\int \frac{dx}{x^2 - 9} =\)
- A \(\frac{1}{6} \log \left| \frac{x-3}{x+3} \right| + C\),where C is constant of integration
- B \(\frac{1}{2} \log \left| \frac{x-3}{x+3} \right| + C\),where C is constant of integration
- C \(\frac{1}{6} \log \left| \frac{x+9}{x-9} \right| + C\),where C is constant of integration
- D \(\frac{1}{6} \log \left| \frac{x+3}{x-3} \right| + C\),where C is constant of integration
Answer & Solution
Correct Answer
(A) \(\frac{1}{6} \log \left| \frac{x-3}{x+3} \right| + C\),where C is constant of integration
Step-by-step Solution
Detailed explanation
\( \int \frac{dx}{x^2 - 9} = \int \frac{dx}{x^2 - 3^2} \) \( = \frac{1}{2 \cdot 3} \log \left| \frac{x-3}{x+3} \right| + C \) \( = \frac{1}{6} \log \left| \frac{x-3}{x+3} \right| + C \)
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