1. **State the problem:** We want to evaluate the integral $$I = \int \frac{x \sin \sqrt{x^2 + 1}}{\sqrt{x^2 + 1}} \, dx.$$\n\n2. **Identify substitution:** Let $$u = \sqrt{x^2 + 1} = (x^2 + 1)^{1/2}.$$\nThen, $$u^2 = x^2 + 1 \implies 2u \frac{du}{dx} = 2x \implies \frac{du}{dx} = \frac{x}{u}.$$\nSo, $$du = \frac{x}{u} dx \implies x dx = u du.$$\n\n3. **Rewrite the integral:** Substitute into the integral:\n$$I = \int \frac{x \sin u}{u} dx = \int \sin u \cdot \frac{x}{u} dx = \int \sin u \, du$$ because $$\frac{x}{u} dx = du.$$\n\n4. **Integrate:** The integral becomes $$\int \sin u \, du = -\cos u + C.$$\n\n5. **Back-substitute:** Replace $$u = \sqrt{x^2 + 1}$$ to get the final answer:\n$$I = -\cos \left( \sqrt{x^2 + 1} \right) + C.$$