Question

The value of \( \int_{0}^{1} \frac{dx}{e^x + e} \) is equal to:

• A. \( \frac{1}{e} \log \left( \frac{1 + e}{2} \right) \)
• B. \( \log \left( \frac{1 + e}{2} \right) \)
• C. \( \frac{1}{e} \log (1 + e) \)
• D. \( \log \left( \frac{2}{1 + e} \right) \)
• E. \( \frac{1}{e} \log \left( \frac{2}{1 + e} \right) \)

Hint:

Whenever you see expressions like \(e^x\) and \(e^{-x}\) together, multiplying or dividing by \(e^x\) or \(e^{-x}\) often converts the integral into a simple logarithmic form.

Answer 1

The Correct Option is A

Concept: We simplify the integrand by dividing numerator and denominator by \( e^x \), which helps convert the expression into a standard logarithmic form suitable for substitution.

Step 1: Rewrite the integrand in a simpler form.
Given: \[ \int_{0}^{1} \frac{e^{-x}}{1 + e \cdot e^{-x}} \, dx \] Divide numerator and denominator by \(e^x\): \[ e^{-x} = \frac{1}{e^x} \] So the integral becomes: \[ \int_{0}^{1} \frac{\frac{1}{e^x}}{1 + \frac{e}{e^x}} \, dx \] Simplify: \[ = \int_{0}^{1} \frac{e^{-x}}{1 + e e^{-x}} \, dx \]

Step 2: Use substitution.
Let: \[ u = 1 + e \cdot e^{-x} \] Differentiate: \[ du = -e \cdot e^{-x} \, dx \] So, \[ e^{-x} \, dx = -\frac{1}{e} \, du \] Now change limits: When \(x = 0\): \[ u = 1 + e \cdot 1 = 1 + e \] When \(x = 1\): \[ u = 1 + e \cdot e^{-1} = 1 + 1 = 2 \]

Step 3: Evaluate the integral.
Substitute into the integral: \[ \int_{0}^{1} \frac{e^{-x}}{1 + e e^{-x}} dx = -\frac{1}{e} \int_{1+e}^{2} \frac{1}{u} \, du \] Now integrate: \[ = -\frac{1}{e} [\ln u]_{1+e}^{2} \] \[ = -\frac{1}{e} (\ln 2 - \ln(1+e)) \] \[ = \frac{1}{e} (\ln(1+e) - \ln 2) \] \[ = \frac{1}{e} \ln\left(\frac{1+e}{2}\right) \] Final Answer: \[ \boxed{\frac{1}{e} \ln\left(\frac{1+e}{2}\right)} \]