$\int \frac{e^{2x} - e^{-2x}}{e^{2x} + e^{-2x}}dx$ is equal to

Given: $\int \frac{e^{2x} - e^{-2x}}{e^{2x} + e^{-2x}}dx$

The numerator is almost the derivative of the denominator.

Let $u = e^{2x} + e^{-2x}$

Differentiate $u$ with respect to $x$:

$\frac{du}{dx} = \frac{d}{dx}(e^{2x}) + \frac{d}{dx}(e^{-2x})$

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

$\frac{du}{dx} = 2(e^{2x} - e^{-2x})$

Therefore: $du = 2(e^{2x} - e^{-2x})dx$

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

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

$= \frac{1}{2}\int \frac{1}{u}du$

$= \frac{1}{2}\log_e|u| + C$

Replace $u$ with $e^{2x} + e^{-2x}$:

$= \frac{1}{2}\log_e|e^{2x} + e^{-2x}| + C$

Therefore, $\int \frac{e^{2x} - e^{-2x}}{e^{2x} + e^{-2x}}dx = \frac{1}{2}\log_e|e^{2x} + e^{-2x}| + C$