solve to get a code: \n10\n∫ 2624 / (x² + x + 1) dx\n1\nask other players\ndiscord community

Explanation:

Step1: Complete the square for the denominator

First, complete the square for $x^{2}+x + 1$. We have $x^{2}+x+1=(x+\frac{1}{2})^{2}+\frac{3}{4}$. Let $u=x+\frac{1}{2}$, then $du = dx$. When $x = 1$, $u=\frac{3}{2}$; when $x = 10$, $u=\frac{21}{2}$. The integral becomes $\int_{\frac{3}{2}}^{\frac{21}{2}}\frac{2624}{u^{2}+\frac{3}{4}}du$.

Step2: Use the integral formula $\int\frac{1}{a^{2}+x^{2}}dx=\frac{1}{a}\arctan(\frac{x}{a})+C$

We know that $\int\frac{2624}{u^{2}+\frac{3}{4}}du=2624\int\frac{1}{u^{2}+(\frac{\sqrt{3}}{2})^{2}}du$. According to the formula $\int\frac{1}{a^{2}+x^{2}}dx=\frac{1}{a}\arctan(\frac{x}{a})+C$, here $a = \frac{\sqrt{3}}{2}$, so $2624\int\frac{1}{u^{2}+(\frac{\sqrt{3}}{2})^{2}}du=\frac{2624}{\frac{\sqrt{3}}{2}}\arctan(\frac{u}{\frac{\sqrt{3}}{2}})+C=\frac{5248}{\sqrt{3}}\arctan(\frac{2u}{\sqrt{3}})+C$.

Step3: Evaluate the definite - integral

$\left[\frac{5248}{\sqrt{3}}\arctan(\frac{2u}{\sqrt{3}})\right]_{\frac{3}{2}}^{\frac{21}{2}}=\frac{5248}{\sqrt{3}}\left[\arctan(\frac{2\times\frac{21}{2}}{\sqrt{3}})-\arctan(\frac{2\times\frac{3}{2}}{\sqrt{3}})\right]=\frac{5248}{\sqrt{3}}\left[\arctan(7\sqrt{3})-\arctan(\sqrt{3})\right]$. Using the formula $\arctan A-\arctan B=\arctan\left(\frac{A - B}{1+AB}\right)$, we have $\arctan(7\sqrt{3})-\arctan(\sqrt{3})=\arctan\left(\frac{7\sqrt{3}-\sqrt{3}}{1 + 7\sqrt{3}\times\sqrt{3}}\right)=\arctan\left(\frac{6\sqrt{3}}{1 + 21}\right)=\arctan\left(\frac{6\sqrt{3}}{22}\right)=\arctan\left(\frac{3\sqrt{3}}{11}\right)$. So the value of the integral is $\frac{5248}{\sqrt{3}}\arctan\left(\frac{3\sqrt{3}}{11}\right)\approx\frac{5248}{\sqrt{3}}\times0.437\approx\frac{5248\times0.437}{\sqrt{3}}\approx\frac{2293.376}{\sqrt{3}}\approx1324$.

Answer:

$1324$