1. **State the problem:** We need to evaluate the definite integral $$\int_1^2 x^3 \ln x \, dx$$ and express it in the form $$a c^3 + b$$ where $a$ and $b$ are rational constants. 2. **Recall the integration by parts formula:** $$\int u \, dv = uv - \int v \, du$$ 3. **Choose parts:** Let $$u = \ln x \implies du = \frac{1}{x} dx$$ $$dv = x^3 dx \implies v = \frac{x^4}{4}$$ 4. **Apply integration by parts:** $$\int_1^2 x^3 \ln x \, dx = \left. \frac{x^4}{4} \ln x \right|_1^2 - \int_1^2 \frac{x^4}{4} \cdot \frac{1}{x} dx = \left. \frac{x^4}{4} \ln x \right|_1^2 - \frac{1}{4} \int_1^2 x^3 dx$$ 5. **Evaluate the boundary term:** $$\left. \frac{x^4}{4} \ln x \right|_1^2 = \frac{2^4}{4} \ln 2 - \frac{1^4}{4} \ln 1 = \frac{16}{4} \ln 2 - 0 = 4 \ln 2$$ 6. **Evaluate the remaining integral:** $$\int_1^2 x^3 dx = \left. \frac{x^4}{4} \right|_1^2 = \frac{2^4}{4} - \frac{1^4}{4} = \frac{16}{4} - \frac{1}{4} = \frac{15}{4}$$ 7. **Substitute back:** $$\int_1^2 x^3 \ln x \, dx = 4 \ln 2 - \frac{1}{4} \cdot \frac{15}{4} = 4 \ln 2 - \frac{15}{16}$$ 8. **Express in the form $a c^3 + b$:** Since $c = \ln 2$, and $c^3 = (\ln 2)^3$, but here we have $4 \ln 2$, so the problem likely means $c = \ln 2$ and the expression is linear in $c$, so we write: $$a = 4, \quad b = -\frac{15}{16}$$ **Final answer:** $$\int_1^2 x^3 \ln x \, dx = 4 \ln 2 - \frac{15}{16}$$