CUET · MATHS · PYQ PAPER 2023
\(\int \frac{1}{(x+1)(x+2)} d x=\)
- A \(\log ((x+1)(x+2))+C\); where \(C\) is an arbitrary constant of integration
- B \(\log \left(\frac{x+1}{x+2}\right)+C\); where \(C\) is an arbitrary constant of integration
- C \(\log \left(\frac{x+2}{x+1}\right)+C\); where \(C\) is an arbitrary constant of integration
- D \(-\log ((x+1)(x+2))+C\); where \(C\) is an arbitrary constant of integration
Answer & Solution
Correct Answer
(B) \(\log \left(\frac{x+1}{x+2}\right)+C\); where \(C\) is an arbitrary constant of integration
Step-by-step Solution
Detailed explanation
\(\frac{1}{(x+1)(x+2)} = \frac{1}{x+1} - \frac{1}{x+2}\) \(\int \left(\frac{1}{x+1} - \frac{1}{x+2}\right) d x = \log \left(\frac{x+1}{x+2}\right) + C\)
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