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