CUET · MATHS · PYQ PAPER 2025
\(\int\left(\frac{1}{\log _e t}-\frac{1}{\left(\log _e t\right)^2}\right) d t\) is equal to
- A \(\frac{1}{\log _e t}+C\), where C is constant of integration
- B \(\frac{t}{\log _e t}+C\), where C is constant of integration
- C \(-\frac{t}{\log _e t}+C\), where \(C\) is constant of integration
- D \(-\frac{1}{\log _e t}+C\), where C is constant of integration
Answer & Solution
Correct Answer
(B) \(\frac{t}{\log _e t}+C\), where C is constant of integration
Step-by-step Solution
Detailed explanation
Let \(f(t) = \frac{t}{\log_e t}\). \(f'(t) = \frac{d}{dt}\left(\frac{t}{\log_e t}\right) = \frac{(\log_e t)(1) - t\left(\frac{1}{t}\right)}{(\log_e t)^2}\) \(f'(t) = \frac{\log_e t - 1}{(\log_e t)^2} = \frac{1}{\log_e t} - \frac{1}{(\log_e t)^2}\)…
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