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

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

Let $u = e^{2x}$

Then $\frac{du}{dx} = 2e^{2x} = 2u$

So $dx = \frac{du}{2u}$

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$\int \frac{2dx}{e^{2x}-1} = \int \frac{2}{u-1} \cdot \frac{du}{2u}$

$= \int \frac{du}{u(u-1)}$

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

$1 = A(u-1) + Bu$

Put $u = 0$:

$1 = A(-1)$

$A = -1$

Put $u = 1$:

$1 = B(1)$

$B = 1$

Therefore $\frac{1}{u(u-1)} = \frac{-1}{u} + \frac{1}{u-1}$

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$\int \frac{du}{u(u-1)} = \int \frac{-1}{u}du + \int \frac{1}{u-1}du$

$= -\ln|u| + \ln|u-1| + C$

$= \ln|u-1| - \ln|u| + C$

$= \ln\left|\frac{u-1}{u}\right| + C$

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Substituting $u = e^{2x}$ back:

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

Therefore, the integral equals $\log\left|\frac{e^{2x}-1}{e^{2x}}\right| + C$