1. **State the problem:** Evaluate the integral $$\int \frac{20 \, dx}{(100x^2 + 1)^2}$$. 2. **Rewrite the integral:** Let us set $$a^2 = \frac{1}{100}$$ so that $$a = \frac{1}{10}$$. Then the integral becomes $$\int \frac{20 \, dx}{(100x^2 + 1)^2} = \int \frac{20 \, dx}{\left(\frac{x^2}{a^2} + 1\right)^2}$$. 3. **Use substitution:** Let $$u = 10x$$, so $$x = \frac{u}{10}$$ and $$dx = \frac{du}{10}$$. 4. **Rewrite the integral in terms of $$u$$:** $$\int \frac{20 \, dx}{(100x^2 + 1)^2} = \int \frac{20 \cdot \frac{du}{10}}{(u^2 + 1)^2} = \int \frac{2 \, du}{(u^2 + 1)^2}$$. 5. **Recall the formula for the integral:** $$\int \frac{du}{(u^2 + 1)^2} = \frac{u}{2(u^2 + 1)} + \frac{1}{2} \arctan(u) + C$$. 6. **Apply the formula:** $$\int \frac{2 \, du}{(u^2 + 1)^2} = 2 \times \left( \frac{u}{2(u^2 + 1)} + \frac{1}{2} \arctan(u) \right) + C = \frac{u}{u^2 + 1} + \arctan(u) + C$$. 7. **Substitute back $$u = 10x$$:** $$\int \frac{20 \, dx}{(100x^2 + 1)^2} = \frac{10x}{100x^2 + 1} + \arctan(10x) + C$$. **Final answer:** $$\boxed{\frac{10x}{100x^2 + 1} + \arctan(10x) + C}$$