Discussion
\(\int \frac{e^{x}}{e^{-x} + 3e^{x}}dx =\)
Answer: A
The integral is \(\frac{1}{6}\log_{e}|1 + 3e^{2x}| + C\).
Multiply top and bottom by \(e^{x}\):
\[ \begin{aligned} \int\frac{e^{2x}}{1 + 3e^{2x}}dx \\ &= \frac{1}{6}\int\frac{6e^{2x}}{1 + 3e^{2x}}dx \\ &= \frac{1}{6}\log|1 + 3e^{2x}| + C \end{aligned} \]
because the derivative of \(1 + 3e^{2x}\) is \(6e^{2x}\). \(\frac{1}{3}\) forgets the 2 from \(e^{2x}\), and the options in \(e^{-x} + 3e^{2x}\) do not differentiate back to the integrand.
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