1. **State the problem:** We need to solve the integral $$\int \frac{x^3}{x^4 + 1} \, dx$$. 2. **Identify the formula and approach:** Notice that the denominator is $x^4 + 1$ and the numerator is $x^3$, which is closely related to the derivative of the denominator. 3. **Recall the derivative:** The derivative of the denominator is $$\frac{d}{dx}(x^4 + 1) = 4x^3$$. 4. **Rewrite the integral:** We can express the integral as $$\int \frac{x^3}{x^4 + 1} \, dx = \int \frac{1}{4} \cdot \frac{4x^3}{x^4 + 1} \, dx = \frac{1}{4} \int \frac{4x^3}{x^4 + 1} \, dx$$. 5. **Use substitution:** Let $$u = x^4 + 1$$, then $$du = 4x^3 \, dx$$. 6. **Substitute and integrate:** The integral becomes $$\frac{1}{4} \int \frac{du}{u} = \frac{1}{4} \ln|u| + C = \frac{1}{4} \ln|x^4 + 1| + C$$. 7. **Final answer:** $$\boxed{\int \frac{x^3}{x^4 + 1} \, dx = \frac{1}{4} \ln|x^4 + 1| + C}$$.