The integral $\int\frac{2\, dx}{e^{2x}-1}$ is equal to:

$\log\left|\frac{e^{2x}-1}{e^{2x}}\right|+C$: C is an arbitrary constant

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

Given integral:

$\int \frac{2 \, dx}{e^{2x}-1}$

Substitute $t = e^{2x} \Rightarrow dt = 2 e^{2x} dx \Rightarrow dx = \frac{dt}{2t}$

Then:

$\int \frac{2 dx}{e^{2x}-1} = \int \frac{2 \cdot \frac{dt}{2t}}{t-1} = \int \frac{dt}{t(t-1)}$

Use partial fractions:

$\frac{1}{t(t-1)} = \frac{A}{t} + \frac{B}{t-1} \Rightarrow 1 = A(t-1) + B t$

Let $t=0 \Rightarrow 1 = A(-1) + 0 \Rightarrow A=-1$

Let $t=1 \Rightarrow 1 = 0 + B(1) \Rightarrow B=1$

So: $\frac{1}{t(t-1)} = \frac{-1}{t} + \frac{1}{t-1}$

Integrate:

$\int \frac{1}{t(t-1)} dt = \int \left(\frac{-1}{t} + \frac{1}{t-1}\right) dt = -\ln|t| + \ln|t-1| = \ln\left|\frac{t-1}{t}\right|$

Back-substitute $t = e^{2x}$:

$\int \frac{2 dx}{e^{2x}-1} = \ln\left|\frac{e^{2x}-1}{e^{2x}}\right| + C = \ln|1 - e^{-2x}| + C$

Answer: $\ln\left|\frac{e^{2x}-1}{e^{2x}}\right| + C$