1. **State the problem:** We need to evaluate the definite integral $$\int_1^{10} \frac{2624}{x^2 + x + 1} \, dx.$$\n\n2. **Recall the formula and approach:** The integral involves a rational function with a quadratic denominator. We can complete the square in the denominator to simplify it:\n$$x^2 + x + 1 = \left(x + \frac{1}{2}\right)^2 + \frac{3}{4}.$$\n\n3. **Rewrite the integral:**\n$$\int_1^{10} \frac{2624}{\left(x + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} \, dx.$$\n\n4. **Use substitution:** Let $$u = x + \frac{1}{2},$$ so when $$x=1,$$ $$u=1.5,$$ and when $$x=10,$$ $$u=10.5.$$\n\n5. **Integral becomes:**\n$$\int_{1.5}^{10.5} \frac{2624}{u^2 + \left(\frac{\sqrt{3}}{2}\right)^2} \, du.$$\n\n6. **Recall the integral formula:**\n$$\int \frac{1}{u^2 + a^2} du = \frac{1}{a} \arctan\left(\frac{u}{a}\right) + C,$$ where $$a = \frac{\sqrt{3}}{2}.$$\n\n7. **Apply the formula:**\n$$\int_{1.5}^{10.5} \frac{2624}{u^2 + \left(\frac{\sqrt{3}}{2}\right)^2} du = 2624 \times \frac{1}{\frac{\sqrt{3}}{2}} \left[ \arctan\left( \frac{u}{\frac{\sqrt{3}}{2}} \right) \right]_{1.5}^{10.5}.$$\n\n8. **Simplify the constant factor:**\n$$\frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}}.$$\nSo the integral equals\n$$2624 \times \frac{2}{\sqrt{3}} \left( \arctan\left( \frac{10.5}{\frac{\sqrt{3}}{2}} \right) - \arctan\left( \frac{1.5}{\frac{\sqrt{3}}{2}} \right) \right).$$\n\n9. **Calculate the arguments of arctan:**\n$$\frac{10.5}{\frac{\sqrt{3}}{2}} = 10.5 \times \frac{2}{\sqrt{3}} = \frac{21}{\sqrt{3}},$$\n$$\frac{1.5}{\frac{\sqrt{3}}{2}} = 1.5 \times \frac{2}{\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}.$$\n\n10. **Final expression:**\n$$\int_1^{10} \frac{2624}{x^2 + x + 1} dx = \frac{5248}{\sqrt{3}} \left( \arctan\left( \frac{21}{\sqrt{3}} \right) - \arctan(\sqrt{3}) \right).$$\n\n11. **Evaluate arctan values:**\n$$\arctan(\sqrt{3}) = \frac{\pi}{3}.$$\nThe value $$\arctan\left( \frac{21}{\sqrt{3}} \right)$$ is approximately 1.428 radians.\n\n12. **Numerical approximation:**\n$$\frac{5248}{\sqrt{3}} \approx 3028.68,$$\nso the integral is approximately\n$$3028.68 \times (1.428 - 1.047) = 3028.68 \times 0.381 = 1153.9.$$\n\n**Final answer:**\n$$\boxed{\int_1^{10} \frac{2624}{x^2 + x + 1} dx \approx 1153.9}.$$