1. **State the problem:** Find the primitive (antiderivative) of the function $f(x) = x^3 (x^2 + 1)^{2020}$. 2. **Recall the formula and rules:** To find the antiderivative of a product involving a function and its derivative, substitution is often useful. Here, notice that the inner function is $u = x^2 + 1$. 3. **Substitution:** Let $u = x^2 + 1$. Then, $\frac{du}{dx} = 2x$ or $du = 2x \, dx$. 4. **Rewrite the integral:** $$\int x^3 (x^2 + 1)^{2020} \, dx = \int x^3 u^{2020} \, dx$$ Express $x^3 dx$ in terms of $u$ and $du$: Since $du = 2x \, dx$, then $x \, dx = \frac{du}{2}$. We have $x^3 dx = x^2 (x \, dx) = (x^2) \cdot \frac{du}{2}$. But $x^2 = u - 1$ (from $u = x^2 + 1$). So, $$x^3 dx = (u - 1) \cdot \frac{du}{2} = \frac{u - 1}{2} du$$ 5. **Substitute back into the integral:** $$\int x^3 (x^2 + 1)^{2020} \, dx = \int u^{2020} \cdot \frac{u - 1}{2} du = \frac{1}{2} \int u^{2020} (u - 1) du$$ 6. **Expand the integrand:** $$\frac{1}{2} \int (u^{2021} - u^{2020}) du = \frac{1}{2} \left( \int u^{2021} du - \int u^{2020} du \right)$$ 7. **Integrate each term:** $$\int u^{2021} du = \frac{u^{2022}}{2022}$$ $$\int u^{2020} du = \frac{u^{2021}}{2021}$$ 8. **Combine results:** $$\frac{1}{2} \left( \frac{u^{2022}}{2022} - \frac{u^{2021}}{2021} \right) + C$$ 9. **Substitute back $u = x^2 + 1$:** $$\boxed{\frac{1}{2} \left( \frac{(x^2 + 1)^{2022}}{2022} - \frac{(x^2 + 1)^{2021}}{2021} \right) + C}$$ This is the primitive of the given function.