Question

\( \int \frac{4e^x}{2e^x - 5e^{-x}} dx = \)

• A. \( 4\log|e^x - 5| + C \)
• B. \( \log|2e^x - 5| + C \)
• C. \( \log|2e^x - 5e^{-x}| + C \)
• D. \( 4\log|2e^x - 5| + C \)
• E. \( \log|2e^{2x} - 5| + C \)

Hint:

If denominator has \(e^x\) and \(e^{-x}\), multiply by \(e^x\) to simplify.

Answer 1

Answer: (E)

Concept: Use substitution when numerator resembles derivative of denominator.

Step 1: Simplify expression.
Multiply numerator and denominator by \( e^x \): \[ \frac{4e^x}{2e^x - 5e^{-x}} \cdot \frac{e^x}{e^x} \] \[ = \frac{4e^{2x}}{2e^{2x} - 5} \]

Step 2: Choose substitution.
Let: \[ u = 2e^{2x} - 5 \]

Step 3: Differentiate.
\[ \frac{du}{dx} = 4e^{2x} \] \[ du = 4e^{2x} dx \]

Step 4: Substitute into integral.
\[ \int \frac{4e^{2x}}{2e^{2x} - 5} dx \] \[ = \int \frac{du}{u} \]

Step 5: Integrate.
\[ = \log|u| + C \]

Step 6: Back substitute.
\[ = \log|2e^{2x} - 5| + C \]

Step 7: Final Answer.
\[ \boxed{\log|2e^{2x} - 5| + C} \]