CUET · MATHS · PYQ PAPER 2025
\(\int \frac{(x-1) e^x}{x^2} d x\), \(x > 0\) equals (where C is an arbitrary constant)
- A \(x e^x+C\)
- B \(\frac{e^x}{x}+C\)
- C \(\frac{e^{-x}}{x^2}+C\)
- D \(e^x+\frac{1}{x}+C\)
Answer & Solution
Correct Answer
(B) \(\frac{e^x}{x}+C\)
Step-by-step Solution
Detailed explanation
\(\int \frac{(x-1) e^x}{x^2} d x = \int \frac{d}{dx} \left(\frac{e^x}{x}\right) d x = \frac{e^x}{x} + C\)
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