Step 1: Simplify the integrand

Divide numerator and denominator by $e^{2x}$ to simplify the integrand:

$\int\frac{e^{4x}-1}{e^{4x}+1}dx = \int\frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}dx$

Step 2: Use substitution

Let $u = e^{2x}+e^{-2x}$. Then, $\frac{du}{dx} = 2e^{2x} - 2e^{-2x} = 2(e^{2x}-e^{-2x})$.

So, $du = 2(e^{2x}-e^{-2x})dx$, which means $(e^{2x}-e^{-2x})dx = \frac{1}{2}du$.

Step 3: Rewrite the integral in terms of u

Substituting into the integral, we get:

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

Step 4: Integrate with respect to u

The integral of $\frac{1}{u}$ with respect to $u$ is $\ln|u|$. Therefore,

$\frac{1}{2}\int\frac{1}{u}du = \frac{1}{2}\ln|u| + C$

Step 5: Substitute back for u

Substitute $u = e^{2x}+e^{-2x}$ back into the expression:

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

Since $e^{2x}+e^{-2x}$ is always positive, we can drop the absolute value:

$\frac{1}{2}\ln(e^{2x}+e^{-2x}) + C$