1. **State the problem:** We need to evaluate the integral $$\int \frac{\ln x}{2x^3} \, dx$$. 2. **Rewrite the integral:** Express the integral as $$\int \frac{\ln x}{2x^3} \, dx = \frac{1}{2} \int \frac{\ln x}{x^3} \, dx$$. 3. **Use substitution or integration by parts:** Let’s use integration by parts where we set: - $$u = \ln x$$ so that $$du = \frac{1}{x} dx$$ - $$dv = x^{-3} dx$$ so that $$v = \int x^{-3} dx = \frac{x^{-2}}{-2} = -\frac{1}{2x^2}$$ 4. **Apply integration by parts formula:** $$\int u \, dv = uv - \int v \, du$$ So, $$\int \frac{\ln x}{x^3} dx = \ln x \cdot \left(-\frac{1}{2x^2}\right) - \int \left(-\frac{1}{2x^2}\right) \cdot \frac{1}{x} dx = -\frac{\ln x}{2x^2} + \frac{1}{2} \int x^{-3} dx$$ 5. **Evaluate the remaining integral:** $$\int x^{-3} dx = \frac{x^{-2}}{-2} = -\frac{1}{2x^2}$$ 6. **Substitute back:** $$-\frac{\ln x}{2x^2} + \frac{1}{2} \left(-\frac{1}{2x^2}\right) + C = -\frac{\ln x}{2x^2} - \frac{1}{4x^2} + C$$ 7. **Recall the factor outside the integral:** The original integral was $$\frac{1}{2} \int \frac{\ln x}{x^3} dx$$, so multiply the result by $$\frac{1}{2}$$: $$\frac{1}{2} \left(-\frac{\ln x}{2x^2} - \frac{1}{4x^2}\right) + C = -\frac{\ln x}{4x^2} - \frac{1}{8x^2} + C$$ **Final answer:** $$\int \frac{\ln x}{2x^3} \, dx = -\frac{\ln x}{4x^2} - \frac{1}{8x^2} + C$$