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