Question

The value of $\int_{-1}^{1} \frac{(1 + \sqrt{|x| - x})e^x + (\sqrt{|x| - x})e^{-x}}{e^x + e^{-x}} \, dx$ is equal to

• A. $3 - \frac{2\sqrt{2}}{3}$
• B. $2 + \frac{2\sqrt{2}}{3}$
• C. $1 - \frac{2\sqrt{2}}{3}$
• D. $1 + \frac{2\sqrt{2}}{3}$

Hint:

Simplify the integrand before evaluating the integral.

Answer 1

The Correct Option is D

1. Simplify the integrand: \[ \int_{-1}^{1} \frac{(1 + \sqrt{|x| - x})e^x + (\sqrt{|x| - x})e^{-x}}{e^x + e^{-x}} \, dx \] \[ = \int_{-1}^{1} \frac{(1 + \sqrt{|x| - x})e^x + (\sqrt{|x| - x})e^{-x}}{e^x + e^{-x}} \, dx \] \[ = \int_{-1}^{1} \frac{(1 + \sqrt{|x| - x})e^x + (\sqrt{|x| - x})e^{-x}}{e^x + e^{-x}} \, dx \]
2. Evaluate the integral: \[ = \int_{-1}^{1} (1 + \sqrt{|x| - x}) \, dx \] \[ = \int_{-1}^{1} 1 \, dx + \int_{-1}^{1} \sqrt{|x| - x} \, dx \] \[ = [x]_{-1}^{1} + \int_{0}^{1} \sqrt{x} \, dx \] \[ = 2 + \frac{2\sqrt{2}}{3} \] The result is $2 + \frac{2\sqrt{2}}{3}$.