The given integral is:

$\int_{1}^{10} \frac{2624}{x^2 + x + 1} \, dx$

Solution Process:

1. Analyze the Denominator:
   The denominator is $x^2 + x + 1$. This quadratic does not factorize easily, so we'll consider completing the square: $x^2 + x + 1 = \left(x + \frac{1}{2}\right)^2 + \frac{3}{4}.$

2. Substitution:
   To simplify, substitute: $u = x + \frac{1}{2}, \quad du = dx.$ This shifts the limits of integration: $x = 1 \implies u = 1 + \frac{1}{2} = \frac{3}{2}, \quad x = 10 \implies u = 10 + \frac{1}{2} = \frac{21}{2}.$ The integral becomes: $\int_{\frac{3}{2}}^{\frac{21}{2}} \frac{2624}{u^2 + \frac{3}{4}} \, du.$

3. Factor Adjustment: Factor the denominator to resemble a standard arctangent form: $u^2 + \frac{3}{4} = u^2 + \left(\sqrt{\frac{3}{4}}\right)^2 = u^2 + \left(\frac{\sqrt{3}}{2}\right)^2.$ Rewrite the integral: $\int_{\frac{3}{2}}^{\frac{21}{2}} \frac{2624}{\left(u^2 + \left(\frac{\sqrt{3}}{2}\right)^2\right)} \, du.$

4. Standard Arctangent Form: Use the formula: $\int \frac{1}{x^2 + a^2} \, dx = \frac{1}{a} \arctan\left(\frac{x}{a}\right) + C.$ Here $a = \frac{\sqrt{3}}{2}$. Thus: $\int \frac{2624}{u^2 + \left(\frac{\sqrt{3}}{2}\right)^2} \, du = \frac{2624}{\frac{\sqrt{3}}{2}} \arctan\left(\frac{2u}{\sqrt{3}}\right).$ Simplify the coefficient: $\frac{2624}{\frac{\sqrt{3}}{2}} = \frac{2624 \cdot 2}{\sqrt{3}} = \frac{5248}{\sqrt{3}}.$

5. Apply Limits: Substituting back the limits $u = \frac{3}{2}$ and $u = \frac{21}{2}$: $\frac{5248}{\sqrt{3}} \left[\arctan\left(\frac{2 \cdot \frac{21}{2}}{\sqrt{3}}\right) - \arctan\left(\frac{2 \cdot \frac{3}{2}}{\sqrt{3}}\right)\right].$

6. Final Simplification: Compute: $\arctan\left(\frac{21}{\sqrt{3}}\right) \text{ and } \arctan\left(\frac{3}{\sqrt{3}} = \sqrt{3}\right),$ then substitute them back into the equation for the result.

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