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

$\dfrac{e^{2x} - 1}{e^{2x} + 1}$ can be simplified by dividing both numerator and denominator by $e^x$:

$\dfrac{e^{2x} - 1}{e^{2x} + 1} = \dfrac{\dfrac{e^{2x}}{e^x} - \dfrac{1}{e^x}}{\dfrac{e^{2x}}{e^x} + \dfrac{1}{e^x}} = \dfrac{e^x - e^{-x}}{e^x + e^{-x}}$

Now the numerator is the derivative of the denominator.

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Let $t = e^x + e^{-x}$

$\dfrac{dt}{dx} = e^x - e^{-x}$

$dt = (e^x - e^{-x})\, dx$

The integral becomes:

$\int \dfrac{dt}{t}$

$= \log|t| + C$

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Substituting back $t = e^x + e^{-x}$:

$= \log|e^x + e^{-x}| + C$