Question:
$\int \frac{9e^{x}+4e^{-x}}{9e^{x}-4e^{-x}}dx =$
$\int \frac{9e^{x}+4e^{-x}}{9e^{x}-4e^{-x}}dx =$
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Whenever the numerator is exactly the derivative of the denominator, the answer is always $\log|\text{denominator}|$.
Updated On: Apr 28, 2026
- $9e^{x}-4e^{-x}+C$
- $\log|9e^{x}+4e^{-x}|+C$
- $4e^{x}-9e^{-x}+C$
- $\log|4e^{x}-9e^{-x}|+C$
- $\log|9e^{x}-4e^{-x}|+C$
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The Correct Option is
Solution and Explanation
Step 1: Concept
Observe if the numerator is the derivative of the denominator.
Step 2: Analysis
Let $f(x) = 9e^x - 4e^{-x}$. $f'(x) = 9e^x - 4(e^{-x} \cdot -1) = 9e^x + 4e^{-x}$.
Step 3: Conclusion
The integral is of the form $\int \frac{f'(x)}{f(x)} dx = \log|f(x)| + C$. Result $= \log|9e^x - 4e^{-x}| + C$. Final Answer: (E)
Observe if the numerator is the derivative of the denominator.
Step 2: Analysis
Let $f(x) = 9e^x - 4e^{-x}$. $f'(x) = 9e^x - 4(e^{-x} \cdot -1) = 9e^x + 4e^{-x}$.
Step 3: Conclusion
The integral is of the form $\int \frac{f'(x)}{f(x)} dx = \log|f(x)| + C$. Result $= \log|9e^x - 4e^{-x}| + C$. Final Answer: (E)
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