1. **State the problem:** We need to find the integral $$\int (x+20) \arctan(21x) \, dx$$. 2. **Formula and approach:** Use integration by parts, where $$\int u \, dv = uv - \int v \, du$$. 3. **Choose parts:** Let $$u = \arctan(21x)$$ and $$dv = (x+20) dx$$. 4. **Compute derivatives and integrals:** - $$du = \frac{d}{dx} \arctan(21x) dx = \frac{21}{1+(21x)^2} dx = \frac{21}{1+441x^2} dx$$ - $$v = \int (x+20) dx = \frac{x^2}{2} + 20x$$ 5. **Apply integration by parts:** $$\int (x+20) \arctan(21x) dx = \left(\arctan(21x) \right) \left(\frac{x^2}{2} + 20x\right) - \int \left(\frac{x^2}{2} + 20x\right) \frac{21}{1+441x^2} dx$$ 6. **Simplify the integral:** $$= \left(\frac{x^2}{2} + 20x\right) \arctan(21x) - 21 \int \frac{\frac{x^2}{2} + 20x}{1+441x^2} dx$$ 7. **Rewrite the integral inside:** $$\int \frac{\frac{x^2}{2} + 20x}{1+441x^2} dx = \int \frac{x^2/2}{1+441x^2} dx + \int \frac{20x}{1+441x^2} dx$$ 8. **Evaluate each integral separately:** - For $$\int \frac{x^2/2}{1+441x^2} dx$$, rewrite numerator: $$x^2 = \frac{1+441x^2 - 1}{441}$$ so $$\frac{x^2}{2} = \frac{1}{2} \cdot \frac{1+441x^2 - 1}{441} = \frac{1+441x^2}{882} - \frac{1}{882}$$ Thus, $$\int \frac{x^2/2}{1+441x^2} dx = \int \frac{\frac{1+441x^2}{882} - \frac{1}{882}}{1+441x^2} dx = \int \left(\frac{1}{882} - \frac{1}{882(1+441x^2)}\right) dx = \int \frac{1}{882} dx - \int \frac{1}{882(1+441x^2)} dx$$ - For $$\int \frac{20x}{1+441x^2} dx$$, use substitution: Let $$w = 1 + 441x^2$$, then $$dw = 882x dx$$, so $$x dx = \frac{dw}{882}$$. Rewrite integral: $$\int \frac{20x}{w} dx = 20 \int \frac{x}{w} dx = 20 \int \frac{1}{w} \cdot x dx = 20 \int \frac{1}{w} \cdot \frac{dw}{882} = \frac{20}{882} \int \frac{1}{w} dw = \frac{10}{441} \ln|w| + C = \frac{10}{441} \ln(1+441x^2) + C$$ 9. **Combine all parts:** $$\int \frac{x^2/2 + 20x}{1+441x^2} dx = \int \frac{x^2/2}{1+441x^2} dx + \int \frac{20x}{1+441x^2} dx = \int \frac{1}{882} dx - \int \frac{1}{882(1+441x^2)} dx + \frac{10}{441} \ln(1+441x^2) + C$$ 10. **Evaluate remaining integrals:** - $$\int \frac{1}{882} dx = \frac{x}{882}$$ - $$\int \frac{1}{882(1+441x^2)} dx = \frac{1}{882} \int \frac{1}{1+(21x)^2} dx = \frac{1}{882} \cdot \frac{1}{21} \arctan(21x) = \frac{1}{18522} \arctan(21x)$$ 11. **Final expression:** $$\int (x+20) \arctan(21x) dx = \left(\frac{x^2}{2} + 20x\right) \arctan(21x) - 21 \left( \frac{x}{882} - \frac{1}{18522} \arctan(21x) + \frac{10}{441} \ln(1+441x^2) \right) + C$$ 12. **Simplify coefficients:** - $$21 \cdot \frac{x}{882} = \frac{21x}{882} = \frac{x}{42}$$ - $$21 \cdot \frac{1}{18522} = \frac{21}{18522} = \frac{1}{882}$$ - $$21 \cdot \frac{10}{441} = \frac{210}{441} = \frac{10}{21}$$ So, $$= \left(\frac{x^2}{2} + 20x\right) \arctan(21x) - \frac{x}{42} + \frac{1}{882} \arctan(21x) - \frac{10}{21} \ln(1+441x^2) + C$$ 13. **Group $$\arctan(21x)$$ terms:** $$\left(\frac{x^2}{2} + 20x + \frac{1}{882}\right) \arctan(21x) - \frac{x}{42} - \frac{10}{21} \ln(1+441x^2) + C$$ **Final answer:** $$\boxed{\int (x+20) \arctan(21x) dx = \left(\frac{x^2}{2} + 20x + \frac{1}{882}\right) \arctan(21x) - \frac{x}{42} - \frac{10}{21} \ln(1+441x^2) + C}$$