Integrate the function $\frac{e^{2x}-1}{e^{2x}+1}$.

(N/A) To integrate the function $I = \int \frac{e^{2x}-1}{e^{2x}+1} dx$,we first divide the numerator and the denominator by $e^{x}$:
$I = \int \frac{\frac{e^{2x}-1}{e^{x}}}{\frac{e^{2x}+1}{e^{x}}} dx = \int \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} dx$
Now,let $t = e^{x}+e^{-x}$.
Differentiating both sides with respect to $x$,we get $dt = (e^{x}-e^{-x}) dx$.
Substituting these into the integral,we have:
$I = \int \frac{dt}{t} = \log |t| + C$
Substituting back the value of $t$,we get:
$I = \log |e^{x}+e^{-x}| + C$,where $C$ is an arbitrary constant.