CUET · MATHS · PYQ PAPER 2023
The integral \(\int e^x\left(\log x+\frac{1}{x^2}\right) d x\) equals for some arbitrary constant k
- A \(e^x\left(\log x+\frac{1}{x^2}\right)+k\)
- B \(e^x\left(\log x-\frac{1}{x}\right)+k\)
- C \(e^x\left(\log x-\frac{1}{x^2}\right)+k\)
- D \(e^x\left(\log x+\frac{1}{x}\right)+k\)
Answer & Solution
Correct Answer
(B) \(e^x\left(\log x-\frac{1}{x}\right)+k\)
Step-by-step Solution
Detailed explanation
\(\int e^x\left(\log x+\frac{1}{x^2}\right) d x\) \(= \int e^x\left[\left(\log x - \frac{1}{x}\right) + \frac{d}{dx}\left(\log x - \frac{1}{x}\right)\right] d x\) \(= e^x\left(\log x-\frac{1}{x}\right)+k\)
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