Question

Evaluate the integral: $ \int e^{x + \frac{1}{x}} \frac{x^2 - 1}{x^2} \, dx $

• A. \( e^{x + \frac{1}{x}} \)
• B. \( e^{x + \frac{1}{x}} \left( \frac{x^2 - 1}{x^2} \right) \)
• C. \( \frac{e^{x + \frac{1}{x}}}{x} \)
• D. \( e^{x + \frac{1}{x}} \left( \ln |x| - 1 \right) \)

Hint:

Recognizing that the integrand is the derivative of an exponential function makes the integration much simpler.

Answer 1

The Correct Option is A

We are tasked with evaluating the integral: \[ \int e^{x + \frac{1}{x}} \frac{x^2 - 1}{x^2} \, dx \] First, rewrite the integrand: \[ \frac{x^2 - 1}{x^2} = 1 - \frac{1}{x^2} \] 
Thus, the integral becomes: \[ \int e^{x + \frac{1}{x}} \left( 1 - \frac{1}{x^2} \right) \, dx \] Now, observe that the expression inside the integral simplifies to the derivative of \( e^{x + \frac{1}{x}} \): \[ \frac{d}{dx} \left( e^{x + \frac{1}{x}} \right) = e^{x + \frac{1}{x}} \left( 1 - \frac{1}{x^2} \right) \] Therefore, the integral simplifies to: \[ \int \frac{d}{dx} \left( e^{x + \frac{1}{x}} \right) \, dx \] Which simply gives: \[ e^{x + \frac{1}{x}} + C \] 
Thus, the correct answer is \( e^{x + \frac{1}{x}} \).