CUET · MATHS · PYQ PAPER 2025
The integral \(\int e^x\left(\frac{x-1}{2 x^2}\right)\)dx is equal to
- A \(\frac{e^x}{2 x}+C\), where C is constant of integration.
- B \(\frac{2 e^x}{x}+C\), where C is constant of integration.
- C \(\frac{e^x}{x}+C\), where C is constant of integration.
- D \(\frac{e^x}{4 x}+C\), where C is constant of integration.
Answer & Solution
Correct Answer
(A) \(\frac{e^x}{2 x}+C\), where C is constant of integration.
Step-by-step Solution
Detailed explanation
\(\int e^x\left(\frac{x-1}{2 x^2}\right)dx = \int e^x\left(\frac{x}{2x^2} - \frac{1}{2 x^2}\right)dx\) \(= \int e^x\left(\frac{1}{2x} - \frac{1}{2 x^2}\right)dx\) Using the formula \(\int e^x(f(x)+f'(x))dx = e^xf(x)+C\) with \(f(x) = \frac{1}{2x}\) and…
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