1. **State the problem:** We need to solve the integral $$\int x^3 e^{x^2} \, dx$$. 2. **Identify the method:** This integral suggests using substitution because of the composite function $e^{x^2}$ and the polynomial $x^3$. 3. **Choose substitution:** Let $$u = x^2$$. Then, $$du = 2x \, dx$$ or $$\frac{du}{2} = x \, dx$$. 4. **Rewrite the integral:** Express $x^3 dx$ in terms of $u$ and $du$: $$x^3 dx = x^2 \cdot x \, dx = u \cdot \frac{du}{2} = \frac{u}{2} du$$. 5. **Substitute into the integral:** $$\int x^3 e^{x^2} \, dx = \int \frac{u}{2} e^u \, du = \frac{1}{2} \int u e^u \, du$$. 6. **Integrate by parts:** For $$\int u e^u \, du$$, let: - $$v = u$$, so $$dv = du$$ - $$dw = e^u du$$, so $$w = e^u$$ Using integration by parts formula $$\int v \, dw = vw - \int w \, dv$$: $$\int u e^u \, du = u e^u - \int e^u \, du = u e^u - e^u + C = e^u (u - 1) + C$$. 7. **Substitute back:** $$\frac{1}{2} \int u e^u \, du = \frac{1}{2} e^u (u - 1) + C = \frac{1}{2} e^{x^2} (x^2 - 1) + C$$. **Final answer:** $$\int x^3 e^{x^2} \, dx = \frac{1}{2} e^{x^2} (x^2 - 1) + C$$