1. **State the problem:** Evaluate the integral $$\int \sum_{n=1}^{\infty} \frac{n}{2025 x^n} \, dx.$$\n\n2. **Rewrite the sum inside the integral:** The sum is $$\sum_{n=1}^{\infty} \frac{n}{2025 x^n} = \frac{1}{2025} \sum_{n=1}^{\infty} n x^{-n}.$$\n\n3. **Recognize the series:** The series $$\sum_{n=1}^{\infty} n r^n = \frac{r}{(1-r)^2}$$ for $$|r|<1$$. Here, $$r = \frac{1}{x}$$, so the series converges if $$|\frac{1}{x}| < 1 \Rightarrow |x| > 1.$$\n\n4. **Apply the formula:** Substitute $$r = \frac{1}{x}$$ into the formula:\n$$\sum_{n=1}^{\infty} n x^{-n} = \frac{\frac{1}{x}}{(1 - \frac{1}{x})^2} = \frac{\frac{1}{x}}{\left(\frac{x-1}{x}\right)^2} = \frac{\frac{1}{x}}{\frac{(x-1)^2}{x^2}} = \frac{1}{x} \cdot \frac{x^2}{(x-1)^2} = \frac{x}{(x-1)^2}.$$\n\n5. **Rewrite the integrand:**\n$$\sum_{n=1}^{\infty} \frac{n}{2025 x^n} = \frac{1}{2025} \cdot \frac{x}{(x-1)^2} = \frac{x}{2025 (x-1)^2}.$$\n\n6. **Integral becomes:**\n$$\int \frac{x}{2025 (x-1)^2} \, dx = \frac{1}{2025} \int \frac{x}{(x-1)^2} \, dx.$$\n\n7. **Simplify the integral:** Let $$u = x-1 \Rightarrow x = u+1$$, so\n$$\int \frac{x}{(x-1)^2} dx = \int \frac{u+1}{u^2} du = \int \left( \frac{u}{u^2} + \frac{1}{u^2} \right) du = \int \left( \frac{1}{u} + u^{-2} \right) du.$$\n\n8. **Integrate term-by-term:**\n$$\int \frac{1}{u} du = \ln|u|,$$\n$$\int u^{-2} du = \int u^{-2} du = -u^{-1} = -\frac{1}{u}.$$\n\n9. **Combine results:**\n$$\int \frac{x}{(x-1)^2} dx = \ln|u| - \frac{1}{u} + C = \ln|x-1| - \frac{1}{x-1} + C.$$\n\n10. **Multiply by the constant:**\n$$\int \sum_{n=1}^{\infty} \frac{n}{2025 x^n} dx = \frac{1}{2025} \left( \ln|x-1| - \frac{1}{x-1} \right) + C.$$\n\n**Final answer:**\n$$\boxed{\int \sum_{n=1}^{\infty} \frac{n}{2025 x^n} dx = \frac{\ln|x-1|}{2025} - \frac{1}{2025 (x-1)} + C}.$$