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 use a standard integral formula.\n\n3. **Complete the square:**\n$$x^2 + x + 1 = \left(x + \frac{1}{2}\right)^2 + 1 - \frac{1}{4} = \left(x + \frac{1}{2}\right)^2 + \frac{3}{4}.$$\n\n4. **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\n5. **Use substitution:** Let $$u = x + \frac{1}{2},$$ so $$du = dx.$$ The limits change to:\nWhen $$x=1,$$ $$u=1 + \frac{1}{2} = \frac{3}{2}.$$\nWhen $$x=10,$$ $$u=10 + \frac{1}{2} = \frac{21}{2}.$$\n\n6. **Integral becomes:**\n$$\int_{\frac{3}{2}}^{\frac{21}{2}} \frac{2624}{u^2 + \left(\frac{\sqrt{3}}{2}\right)^2} \, du.$$\n\n7. **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\n8. **Apply the formula:**\n$$\int_{\frac{3}{2}}^{\frac{21}{2}} \frac{2624}{u^2 + \left(\frac{\sqrt{3}}{2}\right)^2} du = 2624 \times \int_{\frac{3}{2}}^{\frac{21}{2}} \frac{1}{u^2 + \left(\frac{\sqrt{3}}{2}\right)^2} du = 2624 \times \frac{1}{\frac{\sqrt{3}}{2}} \left[ \arctan\left(\frac{u}{\frac{\sqrt{3}}{2}}\right) \right]_{\frac{3}{2}}^{\frac{21}{2}}.$$\n\n9. **Simplify the constant factor:**\n$$\frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}}.$$\n\n10. **Evaluate the definite integral:**\n$$= 2624 \times \frac{2}{\sqrt{3}} \left( \arctan\left( \frac{\frac{21}{2}}{\frac{\sqrt{3}}{2}} \right) - \arctan\left( \frac{\frac{3}{2}}{\frac{\sqrt{3}}{2}} \right) \right) = 2624 \times \frac{2}{\sqrt{3}} \left( \arctan\left( \frac{21}{\sqrt{3}} \right) - \arctan\left( \frac{3}{\sqrt{3}} \right) \right).$$\n\n11. **Simplify inside arctan:**\n$$\frac{3}{\sqrt{3}} = \sqrt{3}.$$\n\n12. **Final answer:**\n$$\boxed{\int_1^{10} \frac{2624}{x^2 + x + 1} dx = \frac{5248}{\sqrt{3}} \left( \arctan\left( \frac{21}{\sqrt{3}} \right) - \arctan(\sqrt{3}) \right)}.$$