The integral ∫(2 dx)/(e^2x-1) is equal to:

The correct answer is Option (4) → \(\log\left|\displaystyle\frac{e^{2x}-1}{e^{2x}}\right|+C\): C is an arbitrary constant

Dividing the numerator and the denominator by \(e^{2x}\) makes the numerator the derivative of the denominator:

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

Let

    \(t = 1 - e^{-2x}, \quad dt = 2e^{-2x}\,dx\)

    \( \displaystyle I = \int \frac{dt}{t} = \log|t| + C \)

    \( \displaystyle I = \log\left|1 - e^{-2x}\right| + C \)

    \( \displaystyle I = \log\left|\frac{e^{2x} - 1}{e^{2x}}\right| + C \)