CUET · MATHS · PYQ PAPER 2025
The integral \(I=\int e^x\left(\frac{x-1}{3 x^2}\right) d x\) is equal to:
- A \(\frac{1}{3}\left(\frac{x^2}{2}-x\right)+C\), where \(C\) is constant of integration
- B \(\left(\frac{x^2}{2}-x\right) e^n+C\), where \(C\) is constant of integration
- C \(\frac{1}{3 x^2} e^x+C\), where \(C\) is constant of integration
- D \(\frac{1}{3 x} e^x+C\), where \(C\) is constant of integration
Answer & Solution
Correct Answer
(D) \(\frac{1}{3 x} e^x+C\), where \(C\) is constant of integration
Step-by-step Solution
Detailed explanation
\(I=\int e^x\left(\frac{x}{3 x^2}-\frac{1}{3 x^2}\right) d x\) \(I=\int e^x\left(\frac{1}{3 x}-\frac{1}{3 x^2}\right) d x\) Let \(f(x) = \frac{1}{3x}\). Then \(f'(x) = -\frac{1}{3x^2}\). Using the formula \(\int e^x(f(x)+f'(x))dx = e^x f(x) + C\).…
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