1. **Stating the problem:** We need to evaluate the integral $$\int \frac{y}{\sqrt{y^2 + 4}} \, dy$$. 2. **Formula and substitution:** A useful substitution for integrals involving expressions like $$\sqrt{y^2 + a^2}$$ is to let $$u = y^2 + 4$$. Then, $$du = 2y \, dy$$ or equivalently $$y \, dy = \frac{du}{2}$$. 3. **Rewrite the integral:** Substitute into the integral: $$\int \frac{y}{\sqrt{y^2 + 4}} \, dy = \int \frac{y}{\sqrt{u}} \, dy$$ Using $$y \, dy = \frac{du}{2}$$, this becomes: $$\int \frac{1}{\sqrt{u}} \cdot \frac{du}{2} = \frac{1}{2} \int u^{-\frac{1}{2}} \, du$$. 4. **Integrate:** Recall the power rule for integration: $$\int u^n \, du = \frac{u^{n+1}}{n+1} + C$$, for $$n \neq -1$$. Here, $$n = -\frac{1}{2}$$, so: $$\frac{1}{2} \int u^{-\frac{1}{2}} \, du = \frac{1}{2} \cdot \frac{u^{\frac{1}{2}}}{\frac{1}{2}} + C = u^{\frac{1}{2}} + C$$. 5. **Back-substitute:** Recall $$u = y^2 + 4$$, so: $$u^{\frac{1}{2}} = \sqrt{y^2 + 4}$$. 6. **Final answer:** $$\int \frac{y}{\sqrt{y^2 + 4}} \, dy = \sqrt{y^2 + 4} + C$$. This completes the solution.