1. **State the problem:** We want to evaluate the integral $$\int \frac{e^{2x}}{(e^{2x} - 1)^3} \, dx$$ where $x \neq 0$. 2. **Identify substitution:** Let $$u = e^{2x} - 1$$ so that $$du = 2e^{2x} dx \implies e^{2x} dx = \frac{du}{2}$$. 3. **Rewrite the integral in terms of $u$:** $$\int \frac{e^{2x}}{(e^{2x} - 1)^3} dx = \int \frac{1}{u^3} \cdot e^{2x} dx = \int \frac{1}{u^3} \cdot \frac{du}{2} = \frac{1}{2} \int u^{-3} du$$. 4. **Integrate using the power rule:** Recall $$\int u^n du = \frac{u^{n+1}}{n+1} + C$$ for $n \neq -1$. Here, $n = -3$, so $$\frac{1}{2} \int u^{-3} du = \frac{1}{2} \cdot \frac{u^{-2}}{-2} + C = -\frac{1}{4} u^{-2} + C$$. 5. **Substitute back to $x$:** $$-\frac{1}{4} (e^{2x} - 1)^{-2} + C$$. **Final answer:** $$\boxed{-\frac{1}{4 (e^{2x} - 1)^2} + C}$$