1. **State the problem:** Find the derivative with respect to $x$ of the function $$f(x) = x^3 \ln\sqrt{x^2 + 1}.$$\n\n2. **Rewrite the function for clarity:** Note that $$\sqrt{x^2 + 1} = (x^2 + 1)^{1/2}.$$ Therefore, \n$$f(x) = x^3 \ln\left((x^2 + 1)^{1/2}\right).$$\n\n3. **Simplify the logarithm using log rules:** \n$$\ln\left((x^2 + 1)^{1/2}\right) = \frac{1}{2} \ln(x^2 + 1).$$\nSo, \n$$f(x) = x^3 \cdot \frac{1}{2} \ln(x^2 + 1) = \frac{x^3}{2} \ln(x^2 + 1).$$\n\n4. **Apply the product rule:** \nThe derivative of a product $u(x)v(x)$ is $$\frac{d}{dx}[uv] = u'v + uv'.$$\nHere, \n$$u = \frac{x^3}{2}, \quad v = \ln(x^2 + 1).$$\n\n5. **Compute derivatives of $u$ and $v$:** \n$$u' = \frac{d}{dx} \left( \frac{x^3}{2} \right) = \frac{3x^2}{2}.$$\n$$v' = \frac{d}{dx} \ln(x^2 + 1) = \frac{1}{x^2 + 1} \cdot 2x = \frac{2x}{x^2 + 1}.$$\n\n6. **Substitute into product rule:** \n$$f'(x) = u'v + uv' = \frac{3x^2}{2} \ln(x^2 + 1) + \frac{x^3}{2} \cdot \frac{2x}{x^2 + 1}.$$\n\n7. **Simplify the second term:** \n$$\frac{x^3}{2} \cdot \frac{2x}{x^2 + 1} = \frac{x^3 \cdot 2x}{2(x^2 + 1)} = \frac{x^4}{x^2 + 1}.$$\n\n8. **Final derivative expression:** \n$$f'(x) = \frac{3x^2}{2} \ln(x^2 + 1) + \frac{x^4}{x^2 + 1}.$$\n\n**Answer:** \n$$\boxed{\frac{d}{dx} \left( x^3 \ln \sqrt{x^2 + 1} \right) = \frac{3x^2}{2} \ln(x^2 + 1) + \frac{x^4}{x^2 + 1}}.$$