1. **State the problem:** Evaluate the indefinite integral $$\int x \arctan(18x) \, dx$$. 2. **Recall the integration by parts formula:** $$\int u \, dv = uv - \int v \, du$$ We choose parts to simplify the integral. Here, let: $$u = \arctan(18x) \quad \Rightarrow \quad du = \frac{18}{1+(18x)^2} dx = \frac{18}{1+324x^2} dx$$ $$dv = x \, dx \quad \Rightarrow \quad v = \frac{x^2}{2}$$ 3. **Apply integration by parts:** $$\int x \arctan(18x) \, dx = \frac{x^2}{2} \arctan(18x) - \int \frac{x^2}{2} \cdot \frac{18}{1+324x^2} dx$$ 4. **Simplify the integral:** $$= \frac{x^2}{2} \arctan(18x) - 9 \int \frac{x^2}{1+324x^2} dx$$ 5. **Rewrite the integrand:** $$\frac{x^2}{1+324x^2} = \frac{1+324x^2 - 1}{1+324x^2} = 1 - \frac{1}{1+324x^2}$$ 6. **Split the integral:** $$\int \frac{x^2}{1+324x^2} dx = \int 1 \, dx - \int \frac{1}{1+324x^2} dx = x - \int \frac{1}{1+(18x)^2} dx$$ 7. **Evaluate the remaining integral:** Recall: $$\int \frac{1}{1+a^2x^2} dx = \frac{1}{a} \arctan(ax) + C$$ Here, $a=18$, so $$\int \frac{1}{1+(18x)^2} dx = \frac{1}{18} \arctan(18x) + C$$ 8. **Substitute back:** $$\int \frac{x^2}{1+324x^2} dx = x - \frac{1}{18} \arctan(18x) + C$$ 9. **Final expression:** $$\int x \arctan(18x) \, dx = \frac{x^2}{2} \arctan(18x) - 9 \left(x - \frac{1}{18} \arctan(18x) \right) + C$$ 10. **Simplify:** $$= \frac{x^2}{2} \arctan(18x) - 9x + \frac{9}{18} \arctan(18x) + C = \frac{x^2}{2} \arctan(18x) - 9x + \frac{1}{2} \arctan(18x) + C$$ **Answer:** $$\boxed{\int x \arctan(18x) \, dx = \frac{x^2}{2} \arctan(18x) - 9x + \frac{1}{2} \arctan(18x) + C}$$