1. **State the problem:** We need to evaluate the integral $$\int \frac{2x^3}{18x + 2x^3} \, dx.$$\n\n2. **Simplify the integrand:** Factor the denominator:\n$$18x + 2x^3 = 2x(9 + x^2).$$\nSo the integrand becomes:\n$$\frac{2x^3}{2x(9 + x^2)} = \frac{x^2}{9 + x^2}.$$\n\n3. **Rewrite the integral:**\n$$\int \frac{x^2}{9 + x^2} \, dx.$$\n\n4. **Use algebraic manipulation:** Write the numerator as $x^2 = (9 + x^2) - 9$ to split the fraction:\n$$\frac{x^2}{9 + x^2} = \frac{9 + x^2 - 9}{9 + x^2} = 1 - \frac{9}{9 + x^2}.$$\n\n5. **Split the integral:**\n$$\int \frac{x^2}{9 + x^2} \, dx = \int 1 \, dx - \int \frac{9}{9 + x^2} \, dx = \int 1 \, dx - 9 \int \frac{1}{9 + x^2} \, dx.$$\n\n6. **Integrate each term:**\n- $$\int 1 \, dx = x.$$\n- For $$\int \frac{1}{9 + x^2} \, dx,$$ use the formula $$\int \frac{1}{a^2 + x^2} \, dx = \frac{1}{a} \arctan\left(\frac{x}{a}\right) + C,$$ where $a = 3$. So,\n$$\int \frac{1}{9 + x^2} \, dx = \frac{1}{3} \arctan\left(\frac{x}{3}\right).$$\n\n7. **Combine results:**\n$$\int \frac{x^2}{9 + x^2} \, dx = x - 9 \cdot \frac{1}{3} \arctan\left(\frac{x}{3}\right) + C = x - 3 \arctan\left(\frac{x}{3}\right) + C.$$\n\n**Final answer:**\n$$\boxed{x - 3 \arctan\left(\frac{x}{3}\right) + C}.$$