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