1. **State the problem:** We need to find the integral $$\int x^5 \ln(x^2 + 2x + 4) \, dx$$. 2. **Formula and approach:** Use integration by parts, where $$\int u \, dv = uv - \int v \, du$$. 3. **Choose parts:** Let $$u = \ln(x^2 + 2x + 4)$$ and $$dv = x^5 \, dx$$. 4. **Compute derivatives and integrals:** - $$du = \frac{2x + 2}{x^2 + 2x + 4} \, dx = \frac{2(x+1)}{x^2 + 2x + 4} \, dx$$ - $$v = \frac{x^6}{6}$$ 5. **Apply integration by parts:** $$\int x^5 \ln(x^2 + 2x + 4) \, dx = \frac{x^6}{6} \ln(x^2 + 2x + 4) - \int \frac{x^6}{6} \cdot \frac{2(x+1)}{x^2 + 2x + 4} \, dx$$ 6. **Simplify the integral:** $$= \frac{x^6}{6} \ln(x^2 + 2x + 4) - \frac{1}{3} \int \frac{x^6 (x+1)}{x^2 + 2x + 4} \, dx$$ 7. **Simplify the integrand:** Multiply numerator: $$x^6 (x+1) = x^7 + x^6$$ 8. **Divide polynomial:** Divide $$x^7 + x^6$$ by $$x^2 + 2x + 4$$: Perform polynomial long division: - Leading term: $$x^7 / x^2 = x^5$$ - Multiply divisor by $$x^5$$: $$x^7 + 2x^6 + 4x^5$$ - Subtract: $$(x^7 + x^6) - (x^7 + 2x^6 + 4x^5) = -x^6 - 4x^5$$ Repeat: - Leading term: $$-x^6 / x^2 = -x^4$$ - Multiply divisor by $$-x^4$$: $$-x^6 - 2x^5 - 4x^4$$ - Subtract: $$(-x^6 - 4x^5) - (-x^6 - 2x^5 - 4x^4) = -2x^5 + 4x^4$$ Repeat: - Leading term: $$-2x^5 / x^2 = -2x^3$$ - Multiply divisor by $$-2x^3$$: $$-2x^5 - 4x^4 - 8x^3$$ - Subtract: $$(-2x^5 + 4x^4) - (-2x^5 - 4x^4 - 8x^3) = 8x^4 + 8x^3$$ Repeat: - Leading term: $$8x^4 / x^2 = 8x^2$$ - Multiply divisor by $$8x^2$$: $$8x^4 + 16x^3 + 32x^2$$ - Subtract: $$(8x^4 + 8x^3) - (8x^4 + 16x^3 + 32x^2) = -8x^3 - 32x^2$$ Repeat: - Leading term: $$-8x^3 / x^2 = -8x$$ - Multiply divisor by $$-8x$$: $$-8x^3 - 16x^2 - 32x$$ - Subtract: $$(-8x^3 - 32x^2) - (-8x^3 - 16x^2 - 32x) = -16x^2 + 32x$$ Repeat: - Leading term: $$-16x^2 / x^2 = -16$$ - Multiply divisor by $$-16$$: $$-16x^2 - 32x - 64$$ - Subtract: $$(-16x^2 + 32x) - (-16x^2 - 32x - 64) = 64x + 64$$ Remainder: $$64x + 64$$ 9. **Rewrite integral:** $$\int \frac{x^7 + x^6}{x^2 + 2x + 4} \, dx = \int (x^5 - x^4 - 2x^3 + 8x^2 - 8x - 16) \, dx + \int \frac{64x + 64}{x^2 + 2x + 4} \, dx$$ 10. **Integrate polynomial part:** $$\int (x^5 - x^4 - 2x^3 + 8x^2 - 8x - 16) \, dx = \frac{x^6}{6} - \frac{x^5}{5} - \frac{x^4}{2} + \frac{8x^3}{3} - 4x^2 - 16x + C$$ 11. **Complete the integral:** $$\int x^5 \ln(x^2 + 2x + 4) \, dx = \frac{x^6}{6} \ln(x^2 + 2x + 4) - \frac{1}{3} \left( \frac{x^6}{6} - \frac{x^5}{5} - \frac{x^4}{2} + \frac{8x^3}{3} - 8x^2 - 16x + \int \frac{64x + 64}{x^2 + 2x + 4} \, dx \right) + C$$ 12. **Simplify constants:** $$= \frac{x^6}{6} \ln(x^2 + 2x + 4) - \frac{x^6}{18} + \frac{x^5}{15} + \frac{x^4}{6} - \frac{8x^3}{9} + \frac{8x^2}{3} + \frac{16x}{3} - \frac{1}{3} \int \frac{64x + 64}{x^2 + 2x + 4} \, dx + C$$ 13. **Integrate the remaining rational function:** Complete the square in denominator: $$x^2 + 2x + 4 = (x+1)^2 + 3$$ Rewrite numerator: $$64x + 64 = 64(x+1)$$ Integral becomes: $$\int \frac{64(x+1)}{(x+1)^2 + 3} \, dx$$ Use substitution $$t = x+1$$: $$\int \frac{64t}{t^2 + 3} \, dt$$ Split integral: $$64 \int \frac{t}{t^2 + 3} \, dt$$ Use substitution $$u = t^2 + 3$$, $$du = 2t dt$$: $$64 \int \frac{t}{u} \frac{du}{2t} = 32 \int \frac{1}{u} \, du = 32 \ln|u| + C = 32 \ln(t^2 + 3) + C$$ 14. **Substitute back:** $$32 \ln((x+1)^2 + 3) = 32 \ln(x^2 + 2x + 4)$$ 15. **Final answer:** $$\int x^5 \ln(x^2 + 2x + 4) \, dx = \frac{x^6}{6} \ln(x^2 + 2x + 4) - \frac{x^6}{18} + \frac{x^5}{15} + \frac{x^4}{6} - \frac{8x^3}{9} + \frac{8x^2}{3} + \frac{16x}{3} - \frac{32}{3} \ln(x^2 + 2x + 4) + C$$