Question

Find: \[ \int \frac{e^{4x} - 1}{e^{4x} + 1} \, dx. \]

Hint:

For rational functions with exponential terms, use substitution to simplify the denominator.

Answer 1

Step 1: Simplify the integrand.
Let: \[ I = \int \frac{e^{4x} - 1}{e^{4x} + 1} \, dx. \] Divide the numerator by the denominator: \[ \frac{e^{4x} - 1}{e^{4x} + 1} = 1 - \frac{2}{e^{4x} + 1}. \] Thus: \[ I = \int \left( 1 - \frac{2}{e^{4x} + 1} \right) \, dx = \int 1 \, dx - 2 \int \frac{1}{e^{4x} + 1} \, dx. \] Step 2: Evaluate the first term.
The integral of 1 is: \[ \int 1 \, dx = x. \] Step 3: Substitute for the second term.
Let \( u = e^{4x} + 1 \), so: \[ du = 4e^{4x} \, dx \quad \Rightarrow \quad \frac{du}{4} = e^{4x} \, dx. \] The integral becomes: \[ \int \frac{1}{e^{4x} + 1} \, dx = \frac{1}{4} \int \frac{1}{u} \, du = \frac{1}{4} \ln|u| + C = \frac{1}{4} \ln|e^{4x} + 1| + C. \] Step 4: Combine the results.
The integral is: \[ I = x - \frac{1}{2} \ln|e^{4x} + 1| + C. \] Conclusion:
The value of the integral is: \[ \boxed{x - \frac{1}{2} \ln|e^{4x} + 1| + C}. \]