1. **State the problem:** We need to find the indefinite integral $$\int \frac{x}{\sqrt{x^2 + 1}} \, dx$$. 2. **Recall the formula and rules:** A useful substitution for integrals involving expressions like $\sqrt{x^2 + a^2}$ is to let $u = x^2 + 1$. Then, $du = 2x \, dx$. 3. **Apply substitution:** Let $$u = x^2 + 1$$ Then $$du = 2x \, dx \implies x \, dx = \frac{du}{2}$$ 4. **Rewrite the integral in terms of $u$:** $$\int \frac{x}{\sqrt{x^2 + 1}} \, dx = \int \frac{x}{\sqrt{u}} \, dx = \int \frac{1}{\sqrt{u}} \cdot x \, dx = \int \frac{1}{\sqrt{u}} \cdot \frac{du}{2} = \frac{1}{2} \int u^{-\frac{1}{2}} \, du$$ 5. **Integrate:** $$\frac{1}{2} \int u^{-\frac{1}{2}} \, du = \frac{1}{2} \cdot \frac{u^{\frac{1}{2}}}{\frac{1}{2}} = u^{\frac{1}{2}} + C$$ 6. **Substitute back to $x$:** $$u^{\frac{1}{2}} + C = \sqrt{x^2 + 1} + C$$ **Final answer:** $$\int \frac{x}{\sqrt{x^2 + 1}} \, dx = \sqrt{x^2 + 1} + C$$