Question

If \[ \int \frac{4e^x + 6e^{-x}}{9e^x - 4e^{-x}} \, dx = Ax + B \log \left( 9e^{2x} - 4 \right) + C, \text{ then } (A, B) = \]

• A. \( \left( \frac{3}{2}, \frac{35}{36} \right) \)
• B. \( \left( \frac{-3}{2}, \frac{-35}{36} \right) \)
• C. \( \left( \frac{-3}{2}, \frac{35}{36} \right) \)
• D. \( \left( \frac{3}{2}, \frac{-35}{36} \right) \)

Hint:

When solving integrals involving exponential functions, try using substitutions that simplify the expression. Look for derivatives of terms in the denominator that match terms in the numerator.

Answer 1

The Correct Option is C

We are given the integral: \[ \int \frac{4e^x + 6e^{-x}}{9e^x - 4e^{-x}} \, dx \] We can start solving this integral by making the substitution: \[ u = 9e^x - 4e^{-x} \] \[ du = (9e^x + 4e^{-x}) \, dx \] This substitution transforms the integral into a form where we can solve for \( A \) and \( B \). By comparing the final expression after integrating with the form given in the problem, we can find that: \[ A = \frac{-3}{2} \quad \text{and} \quad B = \frac{35}{36} \] Thus, \( (A, B) = \left( \frac{-3}{2}, \frac{35}{36} \right) \).