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

Answer:

$2624\left(\frac{2\sqrt{3}\pi}{9}-\frac{2\sqrt{3}\arctan\left(\frac{2 + \sqrt{3}}{3}\right)}{9}\right)$

Explanation:

Step1: Complete the square in denominator

$x^{2}+x + 1=(x+\frac{1}{2})^{2}+\frac{3}{4}$

Step2: Use substitution

Let $u=x+\frac{1}{2}$, $du = dx$. When $x = 1$, $u=\frac{3}{2}$; when $x = 10$, $u=\frac{21}{2}$. The integral becomes $\int_{\frac{3}{2}}^{\frac{21}{2}}\frac{2624}{u^{2}+\frac{3}{4}}du$.

Step3: Rewrite the integral

$\int_{\frac{3}{2}}^{\frac{21}{2}}\frac{2624}{u^{2}+\frac{3}{4}}du=2624\int_{\frac{3}{2}}^{\frac{21}{2}}\frac{1}{u^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}du$

Step4: Use integral formula $\int\frac{1}{a^{2}+x^{2}}dx=\frac{1}{a}\arctan\left(\frac{x}{a}\right)+C$

Here $a=\frac{\sqrt{3}}{2}$, so $2624\int_{\frac{3}{2}}^{\frac{21}{2}}\frac{1}{u^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}du=\frac{2624}{\frac{\sqrt{3}}{2}}\left[\arctan\left(\frac{2u}{\sqrt{3}}\right)\right]_{\frac{3}{2}}^{\frac{21}{2}}$

Step5: Evaluate the definite - integral

$\frac{5248}{\sqrt{3}}\left(\arctan\left(\frac{21}{\sqrt{3}}\right)-\arctan\left(\sqrt{3}\right)\right)=\frac{5248}{\sqrt{3}}\left(\arctan\left(7\sqrt{3}\right)-\frac{\pi}{3}\right)=2624\left(\frac{2\sqrt{3}\pi}{9}-\frac{2\sqrt{3}\arctan\left(\frac{2 + \sqrt{3}}{3}\right)}{9}\right)$