1. **State the problem:** We want to evaluate the integral $$\int \frac{x e^{2x}}{(2x+1)^2} \, dx.$$\n\n2. **Identify the method:** This integral suggests using substitution and integration by parts because of the product of $x e^{2x}$ and the complicated denominator $(2x+1)^2$.\n\n3. **Substitution:** Let $$u = 2x + 1,$$ then $$du = 2 dx \implies dx = \frac{du}{2}.$$ Also, $$x = \frac{u - 1}{2}.$$\n\n4. **Rewrite the integral in terms of $u$:**\n$$\int \frac{x e^{2x}}{(2x+1)^2} dx = \int \frac{\frac{u-1}{2} e^{u-1}}{u^2} \cdot \frac{du}{2} = \int \frac{(u-1) e^{u-1}}{4 u^2} du.$$\n\n5. **Simplify the integral:**\n$$\int \frac{(u-1) e^{u-1}}{4 u^2} du = \frac{1}{4} \int \frac{(u-1) e^{u-1}}{u^2} du.$$\n\n6. **Rewrite the exponential:**\n$$e^{u-1} = e^{-1} e^u,$$ so the integral becomes\n$$\frac{e^{-1}}{4} \int \frac{(u-1) e^u}{u^2} du.$$\n\n7. **Set constant outside:**\n$$\frac{e^{-1}}{4} \int \frac{(u-1) e^u}{u^2} du = \frac{e^{-1}}{4} \int \left(\frac{u}{u^2} - \frac{1}{u^2}\right) e^u du = \frac{e^{-1}}{4} \int \left(\frac{1}{u} - \frac{1}{u^2}\right) e^u du.$$\n\n8. **Split the integral:**\n$$\frac{e^{-1}}{4} \left( \int \frac{e^u}{u} du - \int \frac{e^u}{u^2} du \right).$$\n\n9. **Integration by parts for the second integral:** Let\n$$I_1 = \int \frac{e^u}{u} du, \quad I_2 = \int \frac{e^u}{u^2} du.$$\nWe focus on $I_2$ using integration by parts:\nLet $$v = e^u, \quad dw = \frac{1}{u^2} du,$$ but since $\frac{1}{u^2}$ is not a standard differential, instead use:\n$$I_2 = \int e^u u^{-2} du.$$\nSet:\n$$f = u^{-2}, \quad dg = e^u du,$$ then\n$$df = -2 u^{-3} du, \quad g = e^u.$$\nIntegration by parts formula:\n$$I_2 = f g - \int g df = u^{-2} e^u - \int e^u (-2 u^{-3}) du = \frac{e^u}{u^2} + 2 \int \frac{e^u}{u^3} du.$$\n\n10. **Notice the pattern:** The integral $\int \frac{e^u}{u^n} du$ leads to a recursive pattern increasing the power in the denominator. This integral does not have an elementary closed form and is expressed in terms of the Exponential Integral function $\mathrm{Ei}$ or special functions.\n\n11. **Final expression:** The integral can be expressed as\n$$\int \frac{x e^{2x}}{(2x+1)^2} dx = \frac{e^{-1}}{4} \left( \int \frac{e^u}{u} du - \int \frac{e^u}{u^2} du \right) + C,$$ where $u = 2x + 1$.\n\n12. **Summary:** The integral involves special functions and cannot be simplified further using elementary functions.