CUET · MATHS · PYQ PAPER 2025
\(\int\left(\frac{1}{\log _e x}-\frac{1}{\left(\log _e x\right)^2}\right) d x\) is equal to
- A \(\frac{x}{\log _e x}\) + c: C is an arbitrary constant
- B \(\frac{1}{\log _e x}\) + c : C is an arbitrary constant
- C \(\frac{x}{\left(\log _e x\right)^2}\) + c : C is an arbitrary constant
- D \(\frac{\log _e x}{x}\) + c : C is an arbitrary constant
Answer & Solution
Correct Answer
(A) \(\frac{x}{\log _e x}\) + c: C is an arbitrary constant
Step-by-step Solution
Detailed explanation
Let \(t = \log_e x \Rightarrow x = e^t \Rightarrow dx = e^t dt\) \(\int\left(\frac{1}{t}-\frac{1}{t^2}\right) e^t dt\) Using \(\int e^t (f(t) + f'(t)) dt = e^t f(t) + c\), with \(f(t) = \frac{1}{t}\) \(e^t \cdot \frac{1}{t} + c\) \(x \cdot \frac{1}{\log_e x} + c\)…
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