1. We are asked to evaluate the definite integral $$\int_1^{10} \frac{2624}{x^2 + x + 1} \, dx$$. 2. The integral involves a rational function with a quadratic denominator. To solve it, we first complete the square in the denominator: $$x^2 + x + 1 = \left(x + \frac{1}{2}\right)^2 + \frac{3}{4}$$ 3. Rewrite the integral using this: $$\int_1^{10} \frac{2624}{\left(x + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} \, dx$$ 4. Use the substitution: $$u = x + \frac{1}{2} \implies du = dx$$ 5. Change the limits accordingly: When $x=1$, $u=1 + \frac{1}{2} = \frac{3}{2}$ When $x=10$, $u=10 + \frac{1}{2} = \frac{21}{2}$ 6. The integral becomes: $$\int_{\frac{3}{2}}^{\frac{21}{2}} \frac{2624}{u^2 + \left(\frac{\sqrt{3}}{2}\right)^2} \, du$$ 7. Recall the formula for integrals of the form: $$\int \frac{1}{u^2 + a^2} du = \frac{1}{a} \arctan\left(\frac{u}{a}\right) + C$$ 8. Here, $a = \frac{\sqrt{3}}{2}$, so: $$\int \frac{2624}{u^2 + a^2} du = 2624 \times \frac{1}{a} \arctan\left(\frac{u}{a}\right) + C = \frac{2624}{\frac{\sqrt{3}}{2}} \arctan\left(\frac{2u}{\sqrt{3}}\right) + C$$ 9. Simplify the coefficient: $$\frac{2624}{\frac{\sqrt{3}}{2}} = 2624 \times \frac{2}{\sqrt{3}} = \frac{5248}{\sqrt{3}}$$ 10. Evaluate the definite integral: $$\left. \frac{5248}{\sqrt{3}} \arctan\left(\frac{2u}{\sqrt{3}}\right) \right|_{u=\frac{3}{2}}^{u=\frac{21}{2}} = \frac{5248}{\sqrt{3}} \left[ \arctan\left(\frac{2 \times \frac{21}{2}}{\sqrt{3}}\right) - \arctan\left(\frac{2 \times \frac{3}{2}}{\sqrt{3}}\right) \right]$$ 11. Simplify inside the arctan: $$\arctan\left(\frac{21}{\sqrt{3}}\right) - \arctan\left(\frac{3}{\sqrt{3}}\right) = \arctan\left(\frac{21}{\sqrt{3}}\right) - \arctan\left(\sqrt{3}\right)$$ 12. Note that $\arctan(\sqrt{3}) = \frac{\pi}{3}$. 13. The value $\arctan\left(\frac{21}{\sqrt{3}}\right)$ is approximately $1.446$ radians. 14. Therefore, the integral evaluates to approximately: $$\frac{5248}{\sqrt{3}} (1.446 - 1.047) = \frac{5248}{\sqrt{3}} \times 0.399 \approx 5248 \times 0.230 = 1207.04$$ 15. Final answer: $$\boxed{\int_1^{10} \frac{2624}{x^2 + x + 1} \, dx \approx 1207.04}$$