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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

  1. A \(e^x\left(\log x+\frac{1}{x^2}\right)+k\)
  2. B \(e^x\left(\log x-\frac{1}{x}\right)+k\)
  3. C \(e^x\left(\log x-\frac{1}{x^2}\right)+k\)
  4. D \(e^x\left(\log x+\frac{1}{x}\right)+k\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(e^x\left(\log x-\frac{1}{x}\right)+k\)

Step-by-step Solution

Detailed explanation

\(\int e^x\left(\log x+\frac{1}{x^2}\right) d x\) \(= \int e^x\left[\left(\log x - \frac{1}{x}\right) + \frac{d}{dx}\left(\log x - \frac{1}{x}\right)\right] d x\) \(= e^x\left(\log x-\frac{1}{x}\right)+k\)
From CUET
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