1. **State the problem:** We need to evaluate the integral $$\int \frac{r\sqrt{t} + \sqrt{t}}{t^2} \, dt$$ where $r$ is a constant. 2. **Rewrite the integrand:** Recall that $\sqrt{t} = t^{1/2}$. So the integrand becomes: $$\frac{r t^{1/2} + t^{1/2}}{t^2} = \frac{(r+1) t^{1/2}}{t^2} = (r+1) t^{1/2 - 2} = (r+1) t^{-3/2}$$ 3. **Integral formula:** For any $n \neq -1$, $$\int t^n \, dt = \frac{t^{n+1}}{n+1} + C$$ 4. **Apply the formula:** Here, $n = -\frac{3}{2}$, so $$\int (r+1) t^{-3/2} \, dt = (r+1) \int t^{-3/2} \, dt = (r+1) \frac{t^{-3/2 + 1}}{-3/2 + 1} + C = (r+1) \frac{t^{-1/2}}{-1/2} + C$$ 5. **Simplify the expression:** $$ (r+1) \frac{t^{-1/2}}{-1/2} = (r+1) \times (-2) t^{-1/2} = -2 (r+1) t^{-1/2} + C$$ 6. **Rewrite in radical form:** $$-2 (r+1) \frac{1}{\sqrt{t}} + C = \frac{-2 (r+1)}{\sqrt{t}} + C$$ **Final answer:** $$\int \frac{r\sqrt{t} + \sqrt{t}}{t^2} \, dt = \frac{-2 (r+1)}{\sqrt{t}} + C$$