Question

\(\int\frac{1}{e^{2x}-1}dx=\)

• A. 2log|e^2x-1|-x+C
• B. \(x-\frac{1}{2}\log|e2x-1|+C\)
• C. \(x+\frac{1}{2}\log|e2x-1|+C\)
• D. x-log|e^2x-1|+C
• E. \(\frac{1}{2}\log|e2x-1|-x+C\)

Answer 1

Answer: (E)

We are given the integral \( \int \frac{1}{e^{2x} - 1} \, dx \) and are asked to find the solution. 

We can solve this using substitution. Let:

\( u = e^{2x} - 1 \),

so that \( du = 2e^{2x} \, dx \), or equivalently \( \frac{du}{2} = e^{2x} \, dx \).

Now, rewrite the integral in terms of \( u \):

\( \int \frac{1}{e^{2x} - 1} \, dx = \frac{1}{2} \int \frac{1}{u} \, du \).

The integral of \( \frac{1}{u} \) is \( \log|u| \), so we have:

\( \frac{1}{2} \log|e^{2x} - 1| + C \).

The correct answer is \( \frac{1}{2} \log|e^{2x} - 1| + C \).