Question

If \[ \int \frac{3e^x + 5e^{-x}}{4e^x - 5e^{-x}} \, dx = Ax + B \ln |4e^{2x} - 5| + C \] then, the values of $ A $, $ B $, and $ C $ are:

• A. \( A = -1, B = -\frac{7}{8}, C = \text{constant of integration} \)
• B. \( A = 1, B = \frac{7}{8}, C = \text{constant of integration} \)
• C. \( A = -1, B = \frac{7}{8}, C = \text{constant of integration} \)
• D. \( A = \frac{7}{8}, B = \frac{3}{8}, C = \text{constant of integration} \)

Hint:

For solving integrals of this type, use substitution and differentiation to simplify the equation and identify the constants.

Answer 1

The Correct Option is A

We are given the integral:
\[ I = \int \frac{3e^x + 5e^{-x}}{4e^x - 5e^{-x}} \, dx \] We aim to solve it and express the result in the form: \[ I = Ax + B \ln |4e^{2x} - 5| + C \]
Step 1: Substitution
Let's make a substitution to simplify the integral. Define: \[ u = 4e^x - 5e^{-x} \] Now, differentiate \( u \) with respect to \( x \) to get \( du \): \[ \frac{du}{dx} = 4e^x + 5e^{-x} \] Hence: \[ du = (4e^x + 5e^{-x}) \, dx \] Now, substitute \( du \) and \( u \) into the integral: \[ I = \int \frac{3e^x + 5e^{-x}}{u} \, dx \] 
Step 2: Simplifying the Numerator
Notice that the numerator \( 3e^x + 5e^{-x} \) can be written as: \[ 3e^x + 5e^{-x} = \frac{3}{4}(4e^x + 5e^{-x}) \] Thus, the integral becomes: \[ I = \frac{3}{4} \int \frac{4e^x + 5e^{-x}}{u} \, dx \] From the earlier differentiation, we know that \( 4e^x + 5e^{-x} = du \), so the integral simplifies to: \[ I = \frac{3}{4} \int \frac{du}{u} \] 
Step 3: Integration
The integral of \( \frac{1}{u} \) is \( \ln |u| \), so: \[ I = \frac{3}{4} \ln |u| + C \] 
Step 4: Substituting \( u \) Back
Now, substitute back \( u = 4e^x - 5e^{-x} \) to get: \[ I = \frac{3}{4} \ln |4e^x - 5e^{-x}| + C \] Next, express the result in the desired form by separating the \( x \)-term: \[ I = Ax + B \ln |4e^{2x} - 5| + C \] By comparing both expressions, we find that: \[ A = -1, \quad B = -\frac{7}{8} \]