CUET · MATHS · PYQ PAPER 2025
The value of \(\int\left(\frac{1}{\log _e x}-\frac{1}{\left(\log _e x\right)^2}\right) d x\) is
- A \(x \log _e x+c:\) where \(c\) is an arbitrary constant
- B \(e^{x \log _e x}+c\) : where \(c\) is an arbitrary constant
- C \(\frac{e^x}{\log _e x}+c\) : where \(c\) is an arbitrary constant
- D \(\frac{x}{\log _e x}+c\) : where \(c\) is an arbitrary constant
Answer & Solution
Correct Answer
(D) \(\frac{x}{\log _e x}+c\) : where \(c\) is an arbitrary constant
Step-by-step Solution
Detailed explanation
Let \(t = \log_e x\). Then \(x = e^t\), so \(dx = e^t dt\). \(\int \left(\frac{1}{t} - \frac{1}{t^2}\right) e^t dt\) \(= e^t \cdot \frac{1}{t} + C\) \(= e^{\log_e x} \cdot \frac{1}{\log_e x} + C\) \(= \frac{x}{\log_e x} + C\)
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