Question

$\int \frac{9e^{x}+4e^{-x}}{9e^{x}-4e^{-x}}dx =$

• A. $9e^{x}-4e^{-x}+C$
• B. $\log|9e^{x}+4e^{-x}|+C$
• C. $4e^{x}-9e^{-x}+C$
• D. $\log|4e^{x}-9e^{-x}|+C$
• E. $\log|9e^{x}-4e^{-x}|+C$

Hint:

Whenever the numerator is exactly the derivative of the denominator, the answer is always $\log|\text{denominator}|$.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
We need to find the integral of a rational function involving exponential terms. This integral has a specific form where the numerator is the derivative of the denominator.
Step 2: Key Formula or Approach:
The integral of a function of the form \( \int \frac{f'(x)}{f(x)} dx \) is given by:
\[ \int \frac{f'(x)}{f(x)} dx = \ln|f(x)| + C \] We will check if the given integral fits this pattern.
Step 3: Detailed Explanation:
Let's consider the denominator of the integrand as our function \( f(x) \).
\[ f(x) = 9e^x - 4e^{-x} \] Now, let's find its derivative, \( f'(x) \).
\[ f'(x) = \frac{d}{dx}(9e^x - 4e^{-x}) \] \[ f'(x) = 9 \cdot \frac{d}{dx}(e^x) - 4 \cdot \frac{d}{dx}(e^{-x}) \] \[ f'(x) = 9e^x - 4(e^{-x} \cdot (-1)) \] \[ f'(x) = 9e^x + 4e^{-x} \] We observe that the derivative of the denominator, \( f'(x) \), is exactly equal to the numerator of the integrand.
Therefore, the integral is of the form \( \int \frac{f'(x)}{f(x)} dx \).
Applying the formula, the result is:
\[ \ln|f(x)| + C = \ln|9e^x - 4e^{-x}| + C \] In mathematical texts, \( \log \) often denotes the natural logarithm \( \ln \).
Step 4: Final Answer:
The integral is \( \log|9e^x - 4e^{-x}| + C \).