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:** $$\int u \, dv = uv - \int v \, du$$ We choose: - $u = \ln x$ so that $du = \frac{1}{x} dx$ - $dv = x^3 dx$ so that $v = \frac{x^4}{4}$ 3. **Apply integration by parts:** $$\int_1^2 x^3 \ln x \, dx = \left. \frac{x^4}{4} \ln x \right|_1^2 - \int_1^2 \frac{x^4}{4} \cdot \frac{1}{x} dx = \left. \frac{x^4}{4} \ln x \right|_1^2 - \frac{1}{4} \int_1^2 x^3 dx$$ 4. **Evaluate the remaining integral:** $$\int_1^2 x^3 dx = \left. \frac{x^4}{4} \right|_1^2 = \frac{2^4}{4} - \frac{1^4}{4} = \frac{16}{4} - \frac{1}{4} = \frac{15}{4}$$ 5. **Substitute back:** $$\int_1^2 x^3 \ln x \, dx = \left( \frac{2^4}{4} \ln 2 - \frac{1^4}{4} \ln 1 \right) - \frac{1}{4} \cdot \frac{15}{4} = \left(4 \ln 2 - 0 \right) - \frac{15}{16} = 4 \ln 2 - \frac{15}{16}$$ 6. **Express in the form $a c^3 + b$:** Note that $c = \ln 2$, so: $$\int_1^2 x^3 \ln x \, dx = 4 c - \frac{15}{16}$$ Since the problem states $a c^3 + b$, but our result is linear in $c$, we interpret $a = 0$ and $b = 4 c - \frac{15}{16}$ or more likely the problem expects $a = 4$, $b = -\frac{15}{16}$ with $c = \ln 2$ (and $c^3$ is a typo or misinterpretation). Given the integral, the correct form is: $$a c + b$$ with $a = 4$, $b = -\frac{15}{16}$. **Final answer:** $$\int_1^2 x^3 \ln x \, dx = 4 \ln 2 - \frac{15}{16}$$