Question

Solve \( \int_{-1}^{1} \frac{1+2x}{e^{-x} + e^x} \, dx \).

• A. \( 2 \left( \tan^{-1} e - \frac{\pi}{4} \right) \)
• B. \( 2 \left( \tan^{-1} e - \frac{\pi}{3} \right) \)
• C. \( 2 \left( \tan^{-1} e - \frac{\pi}{2} \right) \)
• D. \( 2 \left( \frac{\pi}{2} - \tan^{-1} e \right) \)

Hint:

When integrating odd functions over symmetric limits, the integral becomes zero.

Answer 1

The Correct Option is A

We are given the integral: \[ I = \int_{-1}^{1} \frac{1+2x}{e^{-x} + e^x} \, dx \] First, recognize that \( e^{-x} + e^x = 2\cosh(x) \), where \( \cosh(x) \) is the hyperbolic cosine function. Thus, the integral becomes: \[ I = \int_{-1}^{1} \frac{1+2x}{2\cosh(x)} \, dx \] Now, split the terms in the numerator: \[ I = \frac{1}{2} \int_{-1}^{1} \frac{1}{\cosh(x)} \, dx + \int_{-1}^{1} \frac{x}{\cosh(x)} \, dx \] The first part of the integral, \( \frac{1}{\cosh(x)} \), is a standard integral, which gives: \[ \int \frac{1}{\cosh(x)} \, dx = \tan^{-1}(\tanh(\frac{x}{2})) \] Evaluating from \(-1\) to \(1\): \[ \int_{-1}^{1} \frac{1}{\cosh(x)} \, dx = 2 \tan^{-1} e \] The second part of the integral involves an odd function \( \frac{x}{\cosh(x)} \), and since we are integrating over a symmetric interval \([-1, 1]\), this part evaluates to zero: \[ \int_{-1}^{1} \frac{x}{\cosh(x)} \, dx = 0 \] Thus, the final answer is: \[ I = 2 \left( \tan^{-1} e - \frac{\pi}{4} \right) \]