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

$\frac{1}{2}\log_e|e^{2x}+e^{-2x}|+C$, where C is an arbitrary constant.

The correct answer is Option (3) → $\frac{1}{2}\log_e|e^{2x}+e^{-2x}|+C$, where C is an arbitrary constant.

$\displaystyle \int \frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}\;dx$

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

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

$\Rightarrow du = 2\left(e^{2x}-e^{-2x}\right)dx$

Thus,

$\displaystyle \int \frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}dx = \frac12 \int \frac{du}{u}$

$= \frac12 \ln|u| + C$

$= \frac12 \ln|e^{2x}+e^{-2x}| + C$

Answer: $\displaystyle \frac12 \ln|e^{2x}+e^{-2x}| + C$