1. **State the problem:** Calculate the definite integral $$\int_0^1 \frac{10}{(x+1)^2} \, dx$$. 2. **Recall the formula:** The integral of $$\frac{1}{(x+a)^n}$$ for $$n \neq 1$$ is $$\int \frac{1}{(x+a)^n} dx = \frac{(x+a)^{1-n}}{1-n} + C$$. 3. **Apply the formula:** Here, $$n=2$$ and $$a=1$$, so $$\int \frac{10}{(x+1)^2} dx = 10 \int (x+1)^{-2} dx = 10 \cdot \frac{(x+1)^{-1}}{-1} + C = -10 (x+1)^{-1} + C = -\frac{10}{x+1} + C$$. 4. **Evaluate the definite integral:** $$\int_0^1 \frac{10}{(x+1)^2} dx = \left[-\frac{10}{x+1}\right]_0^1 = \left(-\frac{10}{2}\right) - \left(-\frac{10}{1}\right) = -5 + 10 = 5$$. 5. **Final answer:** $$\boxed{5}$$