Question

\(\int \frac{e^{x}}{e^{-x} + 3e^{x}} dx =\)

• A. \(\frac{1}{6} \log_{e}|1 + 3e^{2x}| + C\)
• B. \(\log_{e}|e^{-x} + 3e^{2x}| + C\)
• C. \(\frac{1}{3} \log_{e}|1 + 3e^{2x}| + C\)
• D. \(\frac{1}{3} \log_{e}|e^{-x} + 3e^{2x}| + C\)
• E. \(\log_{e}|1 + 3e^{2x}| + C\)

Hint:

When you see \(e^{-x}\) and \(e^{x}\) together, multiply numerator and denominator by \(e^{x}\).

Answer 1

The Correct Option is A

Step 1: Concept:
• Multiply numerator and denominator by \(e^x\).

Step 2: Detailed Explanation:
• Given integral: \[ \int \frac{e^{x}}{e^{-x} + 3e^{x}} \, dx \]
• Multiply numerator and denominator by \(e^x\): \[ = \int \frac{e^{2x}}{1 + 3e^{2x}} \, dx \]
• Let: \[ t = 1 + 3e^{2x} \Rightarrow dt = 6e^{2x} \, dx \Rightarrow e^{2x} \, dx = \frac{dt}{6} \]
• Substitute: \[ \int \frac{e^{2x}}{1 + 3e^{2x}} \, dx = \int \frac{1}{t} \cdot \frac{dt}{6} \]
• Integrate: \[ = \frac{1}{6} \ln|t| + C = \frac{1}{6} \ln|1 + 3e^{2x}| + C \]

Step 3: Final Answer:
• \[ \frac{1}{6} \log_{e}|1 + 3e^{2x}| + C \]