1. **State the problem:** We need to evaluate the integral $$\int \frac{x^2}{4 + x^2} \, dx$$. 2. **Rewrite the integrand:** Notice that $$\frac{x^2}{4 + x^2} = \frac{4 + x^2 - 4}{4 + x^2} = 1 - \frac{4}{4 + x^2}$$. 3. **Split the integral:** Using linearity of integrals, $$\int \frac{x^2}{4 + x^2} \, dx = \int 1 \, dx - \int \frac{4}{4 + x^2} \, dx = \int 1 \, dx - 4 \int \frac{1}{4 + x^2} \, dx$$. 4. **Integrate each term:** - $$\int 1 \, dx = x + C_1$$. - For $$\int \frac{1}{4 + x^2} \, dx$$, recall the formula $$\int \frac{1}{a^2 + x^2} \, dx = \frac{1}{a} \arctan\left(\frac{x}{a}\right) + C$$. Here, $$a = 2$$, so $$\int \frac{1}{4 + x^2} \, dx = \frac{1}{2} \arctan\left(\frac{x}{2}\right) + C_2$$. 5. **Combine results:** $$\int \frac{x^2}{4 + x^2} \, dx = x - 4 \times \frac{1}{2} \arctan\left(\frac{x}{2}\right) + C = x - 2 \arctan\left(\frac{x}{2}\right) + C$$. 6. **Final answer:** $$\boxed{\int \frac{x^2}{4 + x^2} \, dx = x - 2 \arctan\left(\frac{x}{2}\right) + C}$$. This integral was solved by rewriting the integrand to simplify the expression and then applying the standard arctangent integral formula.