1. **State the problem:** Calculate the integral $$\int \frac{x^2}{(x^2+1)^2} \, dx$$. 2. **Recall the formula and rules:** We will use substitution and algebraic manipulation. Important to remember: $$\frac{d}{dx}(x^2+1) = 2x$$ and the integral of rational functions often involves rewriting the numerator. 3. **Rewrite the integrand:** Notice that $$x^2 = (x^2+1) - 1$$, so $$\frac{x^2}{(x^2+1)^2} = \frac{(x^2+1) - 1}{(x^2+1)^2} = \frac{x^2+1}{(x^2+1)^2} - \frac{1}{(x^2+1)^2} = \frac{1}{x^2+1} - \frac{1}{(x^2+1)^2}$$. 4. **Split the integral:** $$\int \frac{x^2}{(x^2+1)^2} \, dx = \int \frac{1}{x^2+1} \, dx - \int \frac{1}{(x^2+1)^2} \, dx$$. 5. **Integrate the first part:** $$\int \frac{1}{x^2+1} \, dx = \arctan(x) + C$$. 6. **Integrate the second part:** Use the known formula: $$\int \frac{1}{(x^2+a^2)^2} \, dx = \frac{x}{2a^2(x^2+a^2)} + \frac{1}{2a^3} \arctan\left(\frac{x}{a}\right) + C$$. For $$a=1$$, $$\int \frac{1}{(x^2+1)^2} \, dx = \frac{x}{2(x^2+1)} + \frac{1}{2} \arctan(x) + C$$. 7. **Combine results:** $$\int \frac{x^2}{(x^2+1)^2} \, dx = \arctan(x) - \left( \frac{x}{2(x^2+1)} + \frac{1}{2} \arctan(x) \right) + C$$ 8. **Simplify:** $$= \arctan(x) - \frac{x}{2(x^2+1)} - \frac{1}{2} \arctan(x) + C = \frac{1}{2} \arctan(x) - \frac{x}{2(x^2+1)} + C$$. **Final answer:** $$\boxed{\int \frac{x^2}{(x^2+1)^2} \, dx = \frac{1}{2} \arctan(x) - \frac{x}{2(x^2+1)} + C}$$