Question

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

Hint:

When encountering rational functions involving exponentials, simplify using substitution to deal with the denominator effectively.

Answer 1

Step 1: Decompose the integrand.
Consider the integral: \[ I = \int \frac{e^{4x} - 1}{e^{4x} + 1} \, dx. \] We can simplify the integrand by rewriting it as: \[ \frac{e^{4x} - 1}{e^{4x} + 1} = 1 - \frac{2}{e^{4x} + 1}. \] Thus, the integral becomes: \[ I = \int \left( 1 - \frac{2}{e^{4x} + 1} \right) \, dx = \int 1 \, dx - 2 \int \frac{1}{e^{4x} + 1} \, dx. \] Step 2: Evaluate the first term.
The first integral is straightforward: \[ \int 1 \, dx = x. \] Step 3: Solve the second integral using substitution.
To solve the second integral, let: \[ u = e^{4x} + 1. \] Then, the derivative of \( u \) is: \[ du = 4e^{4x} \, dx \quad \Rightarrow \quad \frac{du}{4} = e^{4x} \, dx. \] Substitute into the integral: \[ \int \frac{1}{e^{4x} + 1} \, dx = \frac{1}{4} \int \frac{1}{u} \, du = \frac{1}{4} \ln|u| + C = \frac{1}{4} \ln|e^{4x} + 1| + C. \] Step 4: Combine the results.
Now, substitute the results of both integrals: \[ I = x - \frac{1}{2} \ln|e^{4x} + 1| + C. \] Conclusion:
Thus, the value of the integral is: \[ \boxed{x - \frac{1}{2} \ln|e^{4x} + 1| + C}. \]