Discussion
\(\int \frac{(9e^{x} + 4e^{-x})}{(9e^{x} - 4e^{-x})}dx =\)
Answer: E
The integral is \(\log|9e^{x} - 4e^{-x}| + C\).
The numerator is exactly the derivative of the denominator: \(\frac{d}{dx}(9e^{x} - 4e^{-x}) = 9e^{x} + 4e^{-x}\). So the integrand has the form \(\frac{f'(x)}{f(x)}\).
\[ \int \frac{f'(x)}{f(x)}dx = \log|f(x)| + C \]
Using the numerator inside the logarithm, \(\log|9e^{x} + 4e^{-x}|\), would differentiate to the reciprocal of the integrand's ratio. The options without a logarithm, and the one with 4 and 9 swapped, do not differentiate back to the integrand.
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