1. **State the problem:** We are given the integral equation $$\int e^{2x} \left(\ln x + \frac{a}{x}\right) dx = \frac{1}{2} e^{2x} \ln x + c$$ and need to find the value of $a$. 2. **Recall the integration by parts formula:** $$\int u \, dv = uv - \int v \, du$$ 3. **Identify parts:** Let $$u = \ln x \implies du = \frac{1}{x} dx$$ $$dv = e^{2x} dx \implies v = \frac{1}{2} e^{2x}$$ 4. **Apply integration by parts:** $$\int e^{2x} \ln x \, dx = \frac{1}{2} e^{2x} \ln x - \int \frac{1}{2} e^{2x} \frac{1}{x} dx = \frac{1}{2} e^{2x} \ln x - \frac{1}{2} \int \frac{e^{2x}}{x} dx$$ 5. **Given integral:** $$\int e^{2x} \left(\ln x + \frac{a}{x}\right) dx = \int e^{2x} \ln x \, dx + a \int \frac{e^{2x}}{x} dx$$ 6. **Substitute the integration by parts result:** $$= \frac{1}{2} e^{2x} \ln x - \frac{1}{2} \int \frac{e^{2x}}{x} dx + a \int \frac{e^{2x}}{x} dx = \frac{1}{2} e^{2x} \ln x + \left(a - \frac{1}{2}\right) \int \frac{e^{2x}}{x} dx$$ 7. **Compare with given expression:** $$\frac{1}{2} e^{2x} \ln x + c$$ Since the integral expression on the left must equal the right side without any extra integral terms, the coefficient of $$\int \frac{e^{2x}}{x} dx$$ must be zero: $$a - \frac{1}{2} = 0 \implies a = \frac{1}{2}$$ **Final answer:** $$a = \frac{1}{2}$$