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CUET · MATHS · PYQ PAPER 2023

The value of the integral \(\int e^x\left(\frac{1}{x}-\frac{1}{x^2}\right) d x\) is:

  1. A \(\frac{e^x}{x}+C\), where C is a constant
  2. B \(\frac{e^{-x}}{x}+C\), where C is a constant
  3. C \(e^x+C\), where C is a constant
  4. D \(e^{-x}+C\), where C is a constant
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{e^x}{x}+C\), where C is a constant

Step-by-step Solution

Detailed explanation

We use \(\int e^x(f(x)+f'(x))dx = e^x f(x) + C\), where \(f(x)=\frac{1}{x}\) and \(f'(x)=-\frac{1}{x^2}\). \(\int e^x\left(\frac{1}{x}-\frac{1}{x^2}\right) d x = e^x\left(\frac{1}{x}\right)+C\) \(= \frac{e^x}{x}+C\)