1. **State the problem:** We need to evaluate the integral $$\int x^3 \sqrt{1x^2} \, dx$$. 2. **Simplify the integrand:** Since $$\sqrt{1x^2} = \sqrt{x^2} = |x|$$, the integral becomes $$\int x^3 |x| \, dx$$. 3. **Consider the absolute value:** For simplicity, assume $$x \geq 0$$, so $$|x| = x$$. Then the integral is $$\int x^3 \cdot x \, dx = \int x^4 \, dx$$. 4. **Use the power rule for integration:** The formula is $$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$$ for $$n \neq -1$$. 5. **Apply the formula:** Here, $$n=4$$, so $$\int x^4 \, dx = \frac{x^{5}}{5} + C$$. 6. **Final answer:** $$\int x^3 \sqrt{1x^2} \, dx = \frac{x^{5}}{5} + C$$ assuming $$x \geq 0$$. If $$x$$ can be negative, the integral splits into cases due to $$|x|$$, but this is the standard approach for positive $$x$$.