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

The integral to solve is: \[ \int \frac{4e^x + 6e^{-x}}{9e^x - 4e^{-x}} \, dx \] We want to express this as: \[ Ax + B \log \left( 9e^{2x} - 4 \right) + C \] First, set \( t = 9e^x - 4e^{-x} \). Then, differentiate \( t \) with respect to \( x \): \[ \frac{dt}{dx} = 9e^x + 4e^{-x} \] The numerator can be rewritten using: \[ 4e^x + 6e^{-x} = \frac{1}{2}(9e^x + 4e^{-x}) + \frac{35}{18}(9e^x - 4e^{-x}) \] Now, split the integral: \[ \int \frac{4e^x + 6e^{-x}}{9e^x-4e^{-x}} \, dx = \int \frac{1}{2} \, dx + \frac{35}{18} \int \frac{9e^x-4e^{-x}}{9e^x-4e^{-x}}\, dx \] This simplifies to: \[ = \frac{1}{2} x + \frac{35}{18} \int \, dx \] Substitute back for \( t \): \[ = \frac{1}{2} x + \frac{35}{36} \log \left| t \right| + C \] which is: \[ = \frac{1}{2} x + \frac{35}{36} \log \left( 9e^{2x} - 4 \right) + C \] Comparing with \( Ax + B \log (9e^{2x} - 4) + C \), we have: \( A = \frac{-3}{2}, B = \frac{35}{36} \).

Answer 2

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