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CUET · MATHS · PYQ PAPER 2023

\(\int \frac{1}{(x+1)(x+2)} d x=\)

  1. A \(\log ((x+1)(x+2))+C\); where \(C\) is an arbitrary constant of integration
  2. B \(\log \left(\frac{x+1}{x+2}\right)+C\); where \(C\) is an arbitrary constant of integration
  3. C \(\log \left(\frac{x+2}{x+1}\right)+C\); where \(C\) is an arbitrary constant of integration
  4. D \(-\log ((x+1)(x+2))+C\); where \(C\) is an arbitrary constant of integration
Verified Solution

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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